Mathematics 2B study Material( most important questions for 2019-2020 )


			
				
					
						
							
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The Telangana Class 12 Maths Syllabus is designed by the Telangana State Board of Intermediate Education (TSBIE). The board has revised this syllabus by keeping current trends in view. Also, this Intermediate II stage Maths syllabus put emphasis on discovery and understanding of basic mathematical concepts by students.

Telangana Class 12 Maths Syllabus

The Telangana Class 12 Maths Syllabus is divided into 2 parts:

Mathematics II A
Mathematics II B

Telangana Class 12 Mathematics II A Syllabus

The Telangana Mathematics II A Syllabus is basically divided into 2 main topics i.e Algebra & Probability.

ALGEBRA

Chapter 1: Complex Numbers

Chapter 2: De Moivre’s Theorem:

Chapter 3: Quadratic Expressions

Chapter 4: Theory of Equations

Chapter 5: Permutations and Combinations

Chapter 6: Binomial Theorem

Chapter 7: Partial fractions

PROBABILITY

Chapter 8: Measures Of Dispersion

Chapter 9: Probability

Telangana Class 12 Mathematics II B Syllabus

The table below shows the chapter wise distribution of the Telangana Class 12 Maths-II B Syllabus.

CO-ORDINATE GEOMETRY

Chapter 1: Circle

Chapter 2: System of circles

Chapter 3: Parabola

Chapter 4: Ellipse

Chapter 5: Hyperbola

CALCULUS

Chapter 6: Integration

Chapter 7: Definite Integrals

Chapter 8: Differential equation

After knowing the Telangana Class 12 Maths Syllabus it is expected from you that you will bring a shift in your learning process from mere solving the problems in exercises routinely to the conceptual understanding, solving the problems with ingenuity.

Do we hope you find this information on “Telangana Class 12 Maths Syllabus” useful? Stay tuned for more updates on “Telangana Intermediate II stage Exam”. Happy Learning & don’t forget to download the BYJU’S App for interesting videos.

Conics – Introduction

In this section, we shall be studying two of the four non-degenerate conic sections. A conic section is essentially the graph obtained upon slicing a double-ended cone with an infinite plane. Observe the below diagram;

Students should notice that the differing conics all have different angles to the slant of the sides of the cone. This is the factor that determines what shape a conic section.

The four degenerate conic sections are; the parabola, the circle, the ellipse and the hyperbola. Students should already be familiar with the first of the two degenerate conics from the Mathematics course. The last of the two conics will be studied throughout this course.

Conic sections have immense importance in Newton’s theory of Universal Gravitation. All possible trajectories that a satellite may follow are the four non-degenerate conic sections. Physicists and Engineers constantly use these curves to describe possible trajectories of satellites and other celestial objects.

Note that the above information is not required information for the syllabus and that this topic is based on a detailed study of the ellipse and hyperbola, rather than a detailed study of the mechanics of the cone.

Now, all conic sections have a definition in terms of their foci and directrices (plurals for focus and directrix respectively). Those of the circle and parabola should be already known. In general, for any conic section, the definition is;

Let $P$ be a point on the conic section, $S$ and $S'$ be two points called foci. Also, let $M$ and $M'$ be the two points $P$ of intersection of the perpendiculars drawn from the point to two parallel lines called the directrices, with $M$ corresponding to the directrix closer to $S$ and $M'$ corresponding to the directrix closer to $S'$ . The locus of the point $P$ is the conic section required, where $P$ satisifes: $\frac{PS}{PM}=e$

where $e$ is a non-negative constant called the eccentricity.

The value of $e$ determines the curve of the conic section. The above definition of a conic is called the focus-directrix definition, as it involves the foci and directrices in its definition. This is an important definition and must be committed to memory (verbatim).

Cartesian Equations of the ellipse and hyperbola

We shall now study the Cartesian representation of the hyperbola and the ellipse.

The ellipse that is most frequently studied in this course has Cartesian equation;

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

where $a>b>0$ . This form of the ellipse has a graph as shown below.

The points marked $S$ and $S'$ are the two foci of the conic section. The lines marked $M$ and $M'$ are the directrices of the conic. Firstly, some notes on the anatomy of the ellipse. The larger axis of symmetry is dubbed the “major axis” (the line connecting $(-a , 0)$ and $(a , 0)$ ) and the smaller axis of symmetry is called the “minor axis” (the line connecting $(0 , b)$ and $(0 , -b)$ ). Half the length of the major axis is called a “semi major axis” and similarly, half the minor axis is called a “semi minor axis”. Any chord connecting two points with the subsequent line passing through the origin (or centre in general) is called a diameter.

The eccentricity $(e)$ shown before in the focus directrix definition is found to be bounded by $0$ and $1$ . That is, we have that $0<e<1$ , for the ellipse. The formulae defining the eccentricity, foci and directrices are given below for this form of the ellipse.

Component	Formula
Eccentricity $(e)$	$b^{2}=a^{2}(1-e^{2})$
Foci	$(\pm ae , 0)$
Directrices	$x=\pm \frac{a}{e}$

The above formulae must be memorised.

The hyperbola that is most frequently studied in this course has the Cartesian equation;

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Where $a , b>0$ . This form of the hyperbola can be depicted graphically as;

As before, the points marked $S$ and $S'$ are the foci of the conic section, with $M$ and $M'$ representing the directrices. The hyperbola also has another special feature; namely, the existence of asymptotes. As $x\to \pm \infty$ we have that $y\to \pm \frac{b}{a}x$ and these become the oblique asymptotes of the graph. Notice here that the directrices occur closer to the $y$ -axis than the foci do in this case. We shall now study why this is so.

The eccentricity $(e)$ of the hyperbola is bounded below by $1$ and unbounded above. That is the eccentricity of a hyperbola is given by $e>1$ . After observing the formulae for the directrices and foci below, it can be seen that in this case the foci have a larger modulus than do the directrices, since we have that $ae>a$ and that $\frac{a}{e}$ $<a$ .

Component	Formula
Eccentricity	$(e)$ & $b^{2}=a^{2}(e^{2}-1)$
Foci	$(\pm ae , 0)$
Directrices	$x=\pm \frac{a}{e}$
Asymptotes	$y=\pm \frac{b}{a}x$

The above formulae must be memorised.

Note: Although not specifically required by the syllabus, it is interesting to note that the circle has an eccentricity of $0$ and that the parabola has an eccentricity of $1$ .

We shall now look at example questions, looking at drawing graphs of these conic sections.

Example 1

Sketch the graph of $\frac{x^{2}}{9}$ $+$ $\frac{y^{2}}{4}$ $=1$ showing all important features.

Solution 1

Here, “all important features” is equivalent to foci, directrices and intercepts. To check whether we may apply the formulae above, we must firstly have be certain that $a>b$ . In this case $3>2$ and hence $a>b$ .

Now, to find the directrices and foci, we must find the eccentricity. Using the previously shown formulae;

$b^{2}=a^{2}(1-e^{2})$

Upon substitution into the equation we obtain;

$=> 4=9(1-e^{2})$

$e=\frac{\sqrt{5}}{3}$

Now, to obtain the foci we simply substitute into the formula $(\pm ae , 0)$ . Hence we obtain;

$(\pm ae , 0)=(\pm (3)(\frac{\sqrt{5}}{3}), 0)=(\pm \sqrt{5} , 0)$

Now, to obtain the directrices we simply substitute into the formula $x=\pm a/e$ . We then have;

$x=\pm \frac{3}{(\frac{\sqrt{5}}{3})}=\pm \frac{9}{\sqrt{5}}$

Now, we must also find the intercepts. This is easily done, and we obtain $(\pm 3 , 0)$ as the $x$ -intercepts and $(0 , \pm 2)$ as the $y$ -intercepts.

Hence we obtain the graph:

Example 2

Sketch the graph of $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ , showing all important features.

Solution 2

Here, important features imply foci, directrices, asymptotes and intercepts. We firstly find the eccentricity so as to be able to find other features.

$b^{2}=a^{2}(e^{2}-1)$

$9=16(e^{2}-1) e=\frac{5}{4}$

Now, we have that the foci are given by;

$(\pm ae , 0)=(\pm 5 , 0)$

Also, finding the directrices,

$x=\pm \frac{a}{e}=\pm \frac{4}{(\frac{5}{4})}=\pm \frac{16}{5}$

Now, finding the asymptotes;

$y=\pm \frac{b}{a}x=\pm \frac{3}{4}x$

Also, we must find the intercepts with the respective axis, to be able to sketch the required graph. Doing so gives the $x$ -intercepts as $(\pm 4 , 0)$ .

We may now sketch the graph as;

Note: There also exist other forms of the hyperbola and ellipse. The most commonly encountered other form is the ellipse of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ with $b>a$ , and the hyperbola of the form $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1$ . These are each a rotation of the ellipse and hyperbola shown before, by latex]90deg$$ in the counter clockwise direction. We shall not study these as they are very rarely encountered in questions. E.g. no HSC question since the introduction of the new four unit syllabus has asked about conics of the other form. It is sufficient to know the mechanics of the most common form only. For a study of these forms however, see “Cambridge four Unit Mathematics – Conics” by Graham and Denise Arnold.

Parametric Equations of the Ellipse and Hyperbola

We shall now study the parametric equations of the ellipse and hyperbola. At times it becomes useful to represent a conic section in terms of a third variable (in other words a parameter). We shall investigate the parametric representation of the conics, and also look at some of the applications of such representations. We shall not, however look at the derivation of the parametric equations (since this is not required by the syllabus). For information regarding the derivation of the parametric equations, consult “Cambridge Four Unit Mathematics-Conics” by Graham and Denise Arnold.

The ellipse with Cartesian equation $x^{2}/a^{2}+y^{2}/b^{2}=1$ has the parametric representation:

$x=a\cos \theta , y=b\sin \theta$

That is, the point $(x_{1} , y_{1})$ lying on the ellipse $x^{2}/a^{2}$ $+$ $y^{2}/b^{2}$ $=1$ has parametric representation $(a\cos \phi , b\sin \phi )$ where $\phi$ is the value of the parameter at that certain point.

A simple substitution of the parametric equations into the general equation for the ellipse shows that these equations are indeed the parametric equations for the ellipse.

The parametric equations of the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ are given by,

$x=a\sec \theta , y=b\tan \theta$

A simple substitution of the parametric equations into the equation of the hyperbola shows that this is indeed a valid parameterisation. The parametric equations for the ellipse and the hyperbola should be memorised.

At this point, some texts encourage the learning of the equation of the general chord to the hyperbola and ellipse using the parametric representation. However, due to its absence in examination and assessment questions, we shall leave this. If students are interested in seeing a derivation and the equations that form from them, see “Cambridge Four Unit Mathematics – Conics” by Graham and Denise Arnold.

We shall now look at some simple examples illustrating use of the parametric form.

Example 3

Find the parametric equations for the ellipse $x^{2}/16+$ $y^{2}/9$ $=1$ .

Solution 3

Here we have that $a=4$ and $b=3$ . Hence we have that the parametric equations are given by,

$x(\theta )=4\cos \theta , y(\theta )=3\sin \theta$

upon comparison with the formulae given.

Example 4

Find the parametric equations of the hyperbola $x^{2}/12$ $-$ $y^{2}/4$ $=1$

Solution 4

We have that $a=\sqrt{12}=2\sqrt{3}$ and that $b=2$ . Hence we have that the parametric equations of the hyperbola are given by;

$x(\theta )=2\sqrt{3}\sec \theta , y(\theta )=2\tan \theta$

upon comparison with the formulae given.

Example 5

The point $(\sqrt{3} ,\frac{1}{2})$ lies on the ellipse $x^{2}/4$ $+$ $y^{2}=1$ . Find the eccentric angle at this point (i.e. find the value of the parameter at this point).

Solution 5

Firstly we find the parametric equations and then simply equate the coordinates to the parametric equations. Now, we have that $a=2$ and $y=1$ . Hence we have that the parametric equations are

$x=2\cos \theta , y=\sin \theta$

Now, equating these parametric equations to the $x$ and $y$ values that occur at those points, we obtain;

$2\cos \theta =\sqrt{3} , \sin \theta =\frac{1}{2}$

That is,

$\cos \theta =\frac{\sqrt{3}}{2} , \sin \theta =\frac{1}{2}$

Now, the only quadrant where both cosine and sine are positive is in the first quadrant. Hence we have that the eccentric angle is;

$\theta =\frac{\pi }{6}$

Example 6

$P(a\sec \theta , b\tan \theta )$ and $Q(a\sec \phi , b\tan \phi )$ lie on the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ .

a) If $PQ$ subtends a right angle at $(0 , 0)$ , show that $\sin \theta \sin \phi =$ $-$ $\frac{a^{2}}{b^{2}}$ .

b) If $PQ$ subtends a right angle at $(a , 0)$ , show that $\tan (\theta /2)\tan (\phi /2)=-$ $\frac{b^{2}}{a^{2}}$ .

Solution 6

a) We must firstly consider the gradients of $PO$ and $QO$ after which we must consider the product of the gradients which must equal to $-1$ if $PQ$ subtends a right angle at $O(0 ,0)$ .

$m(PO)=\frac{b\tan \theta -0}{a\sec \theta -0}=\frac{b}{a}\sin \theta$

$m(QO)=\frac{b\tan \phi -0}{a\sec \phi -0}=\frac{b}{a}\sin \phi$

Now, since $PQ$ subtends a right angle, then it follows that $POQO$ . That is the product of the gradients must be $-1$ . So we have that,

$m(PO).m(QO)=-1$

=> $(\frac{b}{a}\sin \theta )(\frac{b}{a}\sin \phi )=-1$

$\sin \theta \sin \phi =-\frac{a^{2}}{b^{2}}$

b) Now, we use the same technique as used above, except that we substitute the point $A(a , 0)$ for the point $(0 , 0)$ . So we have that;

$m(PA)=\frac{b\tan \theta -0}{a\sec \theta -a}=\frac{b\tan \theta }{a(\sec \theta -1)}$

and that

$m(QA)=\frac{b\tan \phi -0}{a\sec \phi -a}=\frac{b\tan \phi }{a(\sec \phi -1)}$

Now, we have that

$m(PA).m(QA)=-1$

as before.

=> $(\frac{b\tan \theta }{a(\sec \theta -1)})(\frac{b\tan\phi }{a(\sec \phi -1)})=-1$

=> $\frac{b^{2}\tan \theta \tan \phi }{a^{2}(\sec \theta -1)(\sec \phi -1)}=-1$

Multiplying top and bottom through by $\cos \theta \cos \phi$ gives,

$\frac{b^{2}\sin \theta \sin \phi }{a^{2}(1-\cos \theta )( 1-\cos \phi )}=-1$

Now, we have that $1-\cos x=2\sin ^{2}(x/2)$ . Using this result, and the double angle formulae for sine gives,

$\frac{b^{2}(2\sin \frac{\theta }{2}\cos \frac{\theta }{2})(2\sin \frac{\phi }{2}\cos \frac{\phi }{2})}{a^{2} (2\sin^{2}\frac{\theta }{2})(2\sin ^{2}\frac{\phi }{2})}=-1$

Cancelling out factors and simplifying gives,

$\frac{b^{2}\cos \frac{\theta }{2}\cos \frac{\phi }{2}}{a^{2}\sin \frac{\theta }{2}\sin \frac{\phi }{2}}=-1$

Hence we have;

=> $\cot \frac{\theta }{2}\cot \frac{\phi }{2}=-\frac{a^{2}}{b^{2}}$

Taking the reciprocal of both sides gives,

$\tan \frac{\theta }{2}\tan \frac{\phi }{2}=-\frac{b^{2}}{a^{2}}$

The Tangent, Normal and Chord of Contact to the Ellipse and Hyperbola

The equations of the tangents and normals are not required to be memorised, since most (if not every) question will either ask the student to find any equation to be used or give the student the required equation. If this is not the case, derivation of the equations is quite simple, and consequently remembering the equations is superfluous and not recommended. However, we do recommend that students become familiar with the equations by deriving them as much as possible.

We shall simply list the parametric form and Cartesian form of the required equations.

Tangent

Ellipse

(Cartesian) The equation of the tangent on the ellipse $x^{2}/a^{2}$ $+$ $y^{2}/b^{2}=$ $1$ at the point $P(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}}+\frac{yy_{1}}{b^{2}}=1$

(Parametric) The equation of the tangent at the point $P(a\cos \theta , b\sin \theta )$ on the ellipse $x^{2}/a^{2}$ $+$ $y^{2}/b^{2}=$ $1$ is, $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

Hyperbola

(Cartesian) The equation of the tangent on the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ at the point $P(x_{1} , y_{1})$ is $\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$

(Parametric) The equation of the tangent on the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ at the point $P(a\sec \theta , b\tan \theta )$ is given by, $\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1$

Normal

Ellipse

(Cartesian) The equation of the normal at the point $P(x_{1} , y_{1})$ on the ellipse $x^{2}/a^{2}$ $+$ $y^{2}/b^{2}$ $=1$ is, $\frac{a^{2}x}{x_{1}}-\frac{b^{2}y}{y_{1}}=a^{2}-b^{2}$

(Parametric) The equation of the normal at the point $P(a\cos \theta ,b\sin \theta )$ on the ellipse $x^{2}/a^{2}+$ $y^{2}/b^{2}$ $=1$ is, $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^{2}-b^{2}$

Hyperbola

(Cartesian) The equation of the normal at the point $P(x_{1} , y_{1})$ on the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ is, $\frac{a^{2}x}{x_{1}}+\frac{b^{2}y}{y_{1}}=a^{2}+b^{2}$

(Parametric) The equation of the normal at the point $P(a\sec \theta , b\tan \theta )$ on the hyperbola $x^{2}/a^{2}$ $-$ $y^{2}/b^{2}$ $=1$ is, $\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^{2}+b^{2}$

Note: The derivation of these is quite simple, and students should make it an exercise to find these. A note on differentiating the equations of ellipses and hyperbolas though; differentiate implicitly (recall from the topic graphs) and then simply rearrange to obtain an expression in terms of $x$ and $y$ . After this simply substitute the point into the expression for the first derivative to obtain the gradient at that point. This technique is shown in examples below.

Chord of Contact

The chord of contact from an external point $T(x_{0} , y_{0})$ to a curve, is simply the line connecting the two points of contact of the two tangents drawn from the point $T(x_{0}, y_{0})$ to the curve. Observe the diagram below.

https://aimstutorial.in/wp-content/uploads/chord-of-contact1.png

In this diagram, the chord of contact from the external point $T(x_{0} , y_{0})$ is the line $PQ$ .

The equation of the chord of contact from the external point $T(x_{0} , y_{0})$ on the ellipse $x^{2}/a^{2}$ $+$ $y^{2}/b^{2}$ $=1$ is given by, $\frac{xx_{0}}{a^{2}}+\frac{yy_{0}}{b^{2}}=1$

The equation of the chord of contact from the external point $T(x_{0} , y_{0})$ on the hyperbola $x^{2}/a^{2}-$ $y^{2}/b^{2}$ $=1$ is, $\frac{xx_{0}}{a^{2}}-\frac{yy_{0}}{b^{2}}=1$

The derivation of the chord of contact is carried out in the exact same way that the chord of contact is found for the parabola.

Geometrical Properties of the Hyperbola and Ellipse

There are specific geometrical properties of the hyperbola that are specified by the syllabus. However, these are never directly asked and instead are asked in conjunction with a question. Hence we shall illustrate these properties in questions asking for proofs of them. We shall now look at some example questions to illustrate the nature of the questions asked in this topic.

Example 7

The hyperbola $H$ has equation $9x^{2}-4y^{2}=36$ .

a) Find the coordinates of the foci, $S$ .

b) Find the equations of the directrices.

c) Find the equations of the asymptotes.

d) Sketch the curve $H$ indicating the information obtained in (a) $-$ (c).

e) The point $P(x_{0} , y_{0})$ lies on $H$ . Prove that the equation of the tangent at $P$ is $9xx_{0}-4yy_{0}=36$ .

f) The tangent at $P$ meets the asymptotes at $A$ and $B$ . Prove that $P$ is the midpoint of $AB$ .

Solution 7:

a) To find the coordinates of the foci, we must initially find the eccentricity, after expressing the equation in standard form.

$9x^{2}-4y^{2}=36 \frac{x^{2}}{4}-\frac{y^{2}}{9}=1$

Now, using the equation for the hyperbola gives,

$9=4(e^{2}-1) e=\frac{\sqrt{13}}{2}$

$(\pm \sqrt{13} , 0)$

Hence we have the foci as,

b) The equations of the directrices are given by,

$x=\pm \frac{a}{e}=\frac{2}{(\frac{\sqrt{13}}{2})}=\frac{4}{\sqrt{13}}$

c) The equations of the asymptotes are given by,

$y=\pm \frac{b}{a}x=\pm \frac{3}{2}x$

d) Below is the graph required;

e) So, to find the equation of the tangent, we must simply find the gradient at the point required, and then find the equation by use of the point-gradient formula.

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$

Upon implicitly differentiating with respect to $x$ ,

$\frac{2x}{4}-\frac{2y}{9}(\frac{dy}{dx})=0$

$\frac{dy}{dx}=\frac{9x}{4y}$

At $P$ we have the gradient is equal to,

$\frac{dy}{dx}=\frac{9x_{0}}{4y_{0}}$

Now, at the point $P(x_{0} , y_{0})$ we have that the equation of the tangent is,

$y-y_{0}=\frac{9x_{0}}{4y_{0}}(x-x_{0})$

Upon multiplying through by $4y_{0}$ ,

=> $4yy_{0}-4y_{0}^{ 2}=9xx_{0}-9x_{0}^{ 2}$

That is,

$9xx_{0}-4yy_{0}=9x_{0}^{ 2}-4y_{0}^{ 2}$

=> $9xx_{0}-4yy_{0}=36(\frac{x_{0}^{ 2}}{4}-\frac{y_{0}^{ 2}}{9})$

Now, $P(x_{0} , y_{0})$ lies on the hyperbola and hence we have that;

$\frac{x_{0}^{ 2}}{4}-\frac{y_{0}^{ 2}}{9}=1$

Hence we have,

$9xx_{0}-4yy_{0}=36$

Which is the equation of the tangent at the point $P(x_{0} , y_{0})$ .

f) So, to find the coordinates of $A$ and $B$ we must simply solve simultaneously with the equations of the asymptotes. So consider the asymptote $y=\frac{3}{2}x$ .

$9xx_{0}-4(\frac{3}{2}x)y_{0}=36$

=> $9xx_{0}-6xy_{0}=36$

$x=\frac{12}{3x_{0}-2y_{0}}$

and hence we have that,

$y=\frac{3}{2}(\frac{12}{3x_{0}-2y_{0}})=\frac{18}{3x_{0}-2y_{0}}$

Hence we have that the point $A$ is,

$A(\frac{12}{3x_{0}-2y_{0}} , \frac{18}{3x_{0}-2y_{0}})$

Now, to find the point $B$ we simply solve with the equation $y=-\frac{3}{2}x$ .

$9xx_{0}-4(-\frac{3}{2}x)y_{0}=36$

=> $x=\frac{12}{3x_{0}+2y_{0}}$

=> $y=-\frac{18}{3x_{0}+2y_{0}}$

Hence we have that the point $B$ is given by,

$B(\frac{12}{3x_{0}+2y_{0}},-\frac{18}{3x_{0}+2y_{0}} )$

Now, consider the midpoint of $AB$ .

$M(AB)=(\frac{(\frac{12}{3x_{0}+2y_{0}}+\frac{12}{3x_{0}-2y_{0}})}{2} , \frac{(-\frac{18}{3x_{0}+2y_{0}}+\frac{18}{3x_{0}-2y_{0}})}{2})$

$M(AB)=(6(\frac{3x_{0}-2y_{0}+3x_{0}+2y_{0}}{(3x_{0}+2y_{0})(3x_{0}-2y_{0})}) , 9(\frac{3x_{0}+2y_{0}-(3x_{0}-2y_{0})}{(3x_{0}+2y_{0})(3x_{0}-2y_{0})}))$

=> $M(AB)=(\frac{36x_{0}}{9x_{0}^{ 2}-4y_{0}^{ 2}} , \frac{36y_{0}}{9x_{0}^{ 2}-4y_{0}^{ 2}})$

Now, we have that $P(x_{0} , y_{0})$ lies on the hyperbola, and hence we have that

$\frac{x_{0}^{ 2}}{4}-\frac{y_{0}^{ 2}}{9}=1$

=> $9x_{0}^{2}-4y_{0}^{2}=36$

Thus we have that,

$M(AB)=(\frac{36x_{0}}{36} ,\frac{36y_{0}}{36})=(x_{0} , y_{0})$

Hence we have that,

$M(AB)=P$

That is, the midpoint of $AB$ is the point $P$ .

Example 8

(Reflection property) Prove that the tangent to an ellipse at a point $P$ on it, is equally inclined to the focal chords through $P$ .

Solution 8

Consider the ellipse $\frac{x^{2}}{a^{2}}$ $+$ $\frac{y^{2}}{b^{2}}$ $=1$ and the point $P$ which lies on the ellipse. Consider the points of intersection of the tangents with the directrices being $T$ and $T'$ on the directrices corresponding to the foci $S$ and $S'$ respectively. Also, consider the perpendicular from $P$ to the directrices and the point of contact with the directrices being $M$ and $M'$ corresponding to the foci $S$ and $S'$ respectively. Also, we draw a line through $P$ parallel to the directrices of the ellipse.

Now, the marked angles, namely => $\angle SPT$ and => $\angle S'PT'$ , are the angles we are required to show are equal, since $PS$ and $PS'$ are each focal chords.Now, from the tangent-directrix property (check previous example) the lines $PT$ and $PT$ subtend right angles at their respective foci, namely $S$ and $S'$ respectively. This fact will later become important.Now, the two directrices and the line through $P$ parallel to the directrices are all parallel. By the ratio of intercepts on a transversal theorem (recall from the preliminary Mathematics course) we obtain,

owing to the fact that the three lines form a family of parallel lines.

Now from the focus directrix definition of the ellipse we obtain,

$e=\frac{PS}{PM}=\frac{PS'}{PM'}$

=> $\frac{PM}{PM}=\frac{PS}{PS}$

So, we obtain the expression

$\frac{PT}{PT'}=\frac{PS}{PS'}$

Upon rearranging we obtain

$\frac{PS'}{PT'}=\frac{PS}{PT}$

Now since triangles $PST$ and $PS'T'$ are each right angled, we observe that

=> $\cos (SPT)=\frac{PS}{PT}$

and that

$\cos (S'PT')=\frac{PS'}{PT'}$

Hence we have that

$\cos (SPT)=\cos (S'PT')$

And upon taking the inverse cosines of both sides we obtain

=> $SPT=S'PT'$

since the angles each lie between $0$ and $180deg$ degrees, as they are angles within a triangle. Hence we have that the tangent to an ellipse at a point $P$ on it, is equally inclined to the focal chords through $P$ .

Note: The above proof fuses together algebraic geometry and Euclidean geometry. At times this is the easiest option. Be aware that it can be easier than the alternative use of pure algebra.

The above proof is one of the properties that are required knowledge by the syllabus. Note that there is a corresponding proof for the hyperbola. Students may wish to apply the techniques learnt in the above proof to the hyperbola.

Below is another property required by the syllabus. In this proof we shall investigate the proof or a hyperbola, although it should be noted that there exists a corresponding proof for the ellipse.

Example 9

(Tangent-directrix property) Prove that the part of the tangent between the point of contact and the directrix subtends a right angle at the corresponding focus.

Solution 9

We firstly draw a diagram to depict the question pictorially. Note that we are required to prove that => $PST=90deg$ .

So, consider the hyperbola $\frac{x^{2}}{a^{2}}$ $-$ $\frac{y^{2}}{b^{2}}$ $=1$ and the point $P(x_{1} , y_{1})$ on this hyperbola.We must initially find the coordinates of the point $T$ which is the intersection of the tangent at $P$ with the directrix corresponding to the focu $S(ae , 0)$ . The equation of the tangent at $P$ is given by

$\frac{xx_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}=1$

and hence when $x=$ $\frac{a}{e}$ we have the point of intersection of the tangent with the directrix. Making this substitution gives,

$\frac{(\frac{a}{e})x_{1}}{a^{2}}-\frac{yy_{1}}{b^{2}}=1$

=> $y=\frac{b^{2}(x_{1}-ae)}{y_{1}ae}$

Hence we have that the point $T$ is given by the coordinates,

$T(\frac{a}{e} , \frac{b^{2}(x_{1}-ae)}{y_{1}ae})$

Now, from here we may use any of two methods to prove that => $PST=90deg$ . The first method involves use of Pythagoras’ theorem and consideration of distances. By proving that $\Delta PST$ satisfies Pythagoras’theorem with $PT$ as hypotenuse, we will have successfully proven that => $PST=90deg$ . Another valid method is to use the fact that the product of the gradients of $PS$ and $ST$ must equal $-1$ if => $PST=90deg$ .

We shall use the product of the gradients method as this is by far the easier method. (Note that if one is dealing with numbers, not pronumerals, it is actually easier to use Pythagoras’ theorem rather than the product of the gradients)

So we have that

$m(PS)=\frac{y_{1}-0}{x_{1}-ae}=\frac{y_{1}}{x_{1}-ae}$

$m(ST)=\frac{\frac{b^{2}(x_{1}-ae)}{y_{1}ae}-0}{\frac{a}{e}-ae}=\frac{b^{2}(x_{1}-ae)}{a^{2}y_{1}(1-e^{2})}$

Now, we have that

$m(PS)\times m(ST)=\frac{y_{1}}{x_{1}-ae}\times \frac{b^{2}(x_{1}-ae)}{a^{2}y_{1}(1-e^{2})}$

$m(PS)\times m(ST)=\frac{b^{2}}{a^{2}(1-e^{2})}$

Now, recall that $b^{2}=a^{2}(e^{2}-1)$ . Hence we have that

$m(PS)\times m(ST)=\frac{a^{2}(e^{2}-1)}{a^{2}(1-e^{2})}=-1$

Hence we have that => $PST=90deg$ , and hence this completes the proof.

We shall now look at a further examination style question.

Example 10

a) Show that the equation of the tangent and normal at $P(a\cos \theta , b\sin \theta )$ to the ellipse $\frac{x^{2}}{a^{2}}$ $+$ $\frac{y^{2}}{b^{2}}$ $=1$ are $\frac{x\cos \theta }{a}$ $+$ $\frac{y\sin \theta }{b}$ $=1$ and $\frac{ax}{\cos \theta }$ $-$ $\frac{by}{\sin \theta }$ $=a^{2}-b^{2}$ respectively.

b) The tangent and normal at $P$ cut the $y$ -axis at $A$ and $B$ respectively. Find the coordinates of $A$ and $B$ .

c) Show that the focus $S$ lies on the circumference of the semi-circle which has diameter $AB$ .

Solution 10

a) Differentiating implicitly with respect to $x$ we obtain,

$\frac{2x}{a^{2}}+\frac{2y}{b^{2}} \frac{dy}{dx}=0 \frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}$

Now, substituting in the point $P$ gives,

$\frac{dy}{dx}=-\frac{b\cos \theta }{a\sin \theta }$

So, using the point-gradient formula to find the equation of the tangent at the point $P$ we obtain,

$y-b\sin \theta =-\frac{b\cos \theta }{a\sin \theta }(x-a\cos \theta )$

$a y\sin \theta -ab\sin ^{2}\theta =-bx\cos \theta +ab\cos ^{2}\theta$

=> $bx\cos \theta + ay\sin \theta =ab (\cos ^{2}\theta +\sin ^{2}\theta )$

Hence we have that

$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

Which is the equation of the tangent at the point $P$ . Now, to find the equation of the normal we must find the gradient first, which is the negative reciprocal.

So, the gradient of the normal is given by

$m(N)=\frac{a\sin \theta }{b\cos \theta }$

So, using the point-gradient formula we obtain,

$y-b\sin \theta =\frac{a\sin \theta }{b\cos \theta }(x-a\cos \theta )$

$by\cos \theta -b^{2}\sin \theta \cos \theta =ax\sin \theta -a^{2}\sin \theta \cos \theta$

Upon rearranging ad dividing through by $\sin \theta \cos \theta$ .

=> $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^{2}-b^{2}$

Thus we obtain the equation of the normal.

c) We firstly draw a diagram of this situation.

Firstly, for the tangent at $x=0$ we have the $y$ -intercept which is given by

$y=\frac{b}{\sin \theta }$

For the normal we obtain

$x=0$ we obtain $y=-\frac{\sin \theta }{b}(a^{2}-b^{2})$

Hence we have that $A$ and $B$ are the points given by

$A(0 ,\frac{b}{\sin \theta }) , B(0 , \frac{\sin \theta }{b}(b^{2}-a^{2}))$

To show that the focus $S$ lies on the circumference of the semi-circle which has diameter $AB$ we must show that => $ASB=90deg$ as this shows that $AB$ is the diameter of a circle passing through $S$ by the ‘angle in a semi-circle’ rule in circle geometry.

So we shall do so using gradients.

$(ae , 0)$

$m(AS)=\frac{(\frac{b}{\sin \theta })}{-ae}=-\frac{b}{ae\sin \theta }$

$m(BS)=\frac{\frac{\sin \theta }{b}(b^{2}-a^{2})}{-ae}=-\frac{\sin \theta (b^{2}-a^{2})}{abe}$

$m(AS)\times m(BS)=-\frac{b}{ae\sin \theta }\times -\frac{\sin \theta (b^{2}-a^{2})}{abe}$

$m(AS)\times m(BS)=\frac{b^{2}-a^{2}}{a^{2}e^{2}}$

Now, $b^{2}=a^{2}(1-e^{2})$ . So we obtain,

$m(AS)\times m(BS)=\frac{a^{2}(1-e^{2})-a^{2}}{a^{2}e^{2}}=\frac{a^{2}-a^{2}e^{2}-a^{2}}{a^{2}e^{2}}=-\frac{a^{2}e^{2}}{a^{2}e^{2}}=-1$

Hence => $ASB=90deg$ and thus we have that $S$ lies on the circle with $AB$ as diameter.

Note: Locus problems involving the ellipse and non-rectangular hyperbolae are not included in the syllabus.

The Rectangular Hyperbola

We shall now study a special form of the hyperbola called the rectangular hyperbola. This form of the hyperbola has the special property that the asymptotes are at right angles to each other (Hence the term “rectangular”). Now, considering this and the gradients of the asymptotes we have,

$\frac{b}{a}\times -\frac{b}{a}=-1$

=> $b^{2}=a^{2} b=a$

Since $b , a>0$ . Hence we have that the rectangular hyperbola has equation of the form

$x^{2}-y^{2}=a^{2}$

where $a>0$ . Also, due to the fact that $b=a$ , we thus have that the asymptotes are $y=\pm x$ .

A special property of all rectangular hyperbolae is that the eccentricity is $\sqrt{2}$ . That is $e=\sqrt{2}$ . Here is the proof;

$b^{2}=a^{2}(e^{2}-1)$

Now, recall that $a=b$ for the rectangular hyperbola. Hence we have that

$a^{2}=a^{2}(e^{2}-1)$

That is,

$e=\sqrt{2}$

which is the required result.

Now, the foci, directrices and all other important features regarding the rectangular hyperbola may be found. We shall not present a table, as this encourages memorisation, where as in actual fact, the student should simply apply the above definition of the rectangular hyperbola $(a=b)$ and continue on from there.

The rectangular hyperbola is in general not represented in this form, rather, it is represented in the form $xy=c^{2}$ where $c>0$ . To obtain this form of the rectangular hyperbola, we translate the axes, or in other words, we rotate the curve by $45deg$ in a counter clockwise direction so that the asymptotes become the coordinate axes.

To carry out such a rotation, there are several methods which may be used. One method that may be used is by use of complex numbers. Recall that is a complex number $z=\cos \theta + i\sin \theta$ is multiplied by the complex number $w=\cos \phi +i\sin \phi$ , the resultant complex number is of the form $zw=\cos (\theta +\phi )+i\sin (\theta +\phi )$ . That is the resultant complex number is a rotation of the complex number on the Argand plane by an angle of $\phi$ . We may use this technique to great effect to show that the hyperbola $x^{2}-y^{2}=a^{2}$ converts to the hyperbola $xy=c^{2}$ where $c^{2}=\frac{1}{2}a^{2}$ upon rotation by $45deg$ .

Proof

We are performing the following transformation;

Let $Z=X+iY$ be the complex number to which $z$ is transformed to (consequently, the coordinates to which $z$ is transformed to are $(X , Y)$ ). So we have that,

$Z=(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4})z$

i.e.

$X+iY=(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})(x+iy)$

Expanding out the RHS gives,

$X+iY=(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}})+i(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}})$

Now, equating real and imaginary parts gives,

$X=(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}}) , Y=(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}})$

Now, to eliminate $x$ and $y$ to obtain a new equation that $X$ and $Y$ satisfy, we must multiply the expressions for $X$ and $Y$ .

So, we have

$XY=(\frac{x}{\sqrt{2}}-\frac{y}{\sqrt{2}})(\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}})$

=> $XY=\frac{x^{2}-y^{2}}{2}$

But $x^{2}-y^{2}=a^{2}$ and hence we have that

$XY=\frac{1}{2}a^{2}$

Letting $\frac{1}{2}a^{2}=c^{2}$ we obtain the general form of the equation of the rectangular hyperbola after being rotated by $45deg$ .

$xy=c^{2}$

Now, we may obtain the equations of the directrices and coordinates of the foci by performing similar rotations on each component. We shall not show this, but the results are shown in the table below.

Component	Equation
Foci	$(\pm c\sqrt{2} , \pm c\sqrt{2})$
Directrices	$x+y=\pm c\sqrt{2}$
Vertices	$(\pm c , \pm c)$

The asymptotes now, are obviously $x=0$ and $y=0$ .

Parameterisation of the Rectangular Hyperbola

We shall now consider the parametric representation of the rectangular hyperbola. The rectangular hyperbola $xy=c^{2}$ has parametric representation

$x=ct , y=\frac{c}{t}$

where $t$ is the parameter. It can be seen that upon elimination of the parameter, the equation $xy=c^{2}$ is obtained.

The Tangent, Normal, Chord and Chord of Contact to the Rectangular Hyperbola

We shall now study the equations of the tangent, normal and chord of contact to the rectangular hyperbola. Again, as for the ellipse and hyperbola, you are not required to memorise every formula, instead it is recommended you become familiar with them by solving problems.

Tangent

(Cartesian) The equation of the tangent at the point $(x_{1} , y_{1})$ on the rectangular hyperbola $xy=c^{2}$ is, $xy_{1}+yx_{1}=2c^{2}$

(Parametric) The equation of the tangent at the point $(ct , c/t)$ on the rectangular hyperbola $xy=c^{2}$ is, $x+t^{2}y=2ct$

Normal

(Cartesian) The normal at the point $(x_{1} , y_{1})$ on the rectangular hyperbola $xy=c^{2}$ is given by, $xx_{1}-yy_{1}=x_{1}^{ 2}-y_{1}^{ 2}$

(Parametric) The normal at the point $(ct , c/t)$ on the rectangular hyperbola $xy=c^{2}$ is given by, $tx-\frac{1}{t}y=c(t^{2}-\frac{1}{t^{2}})$

Chord

The equation of the chord joining the points $P(cp , c/p)$ and $Q(cq , c/q)$ on the hyperbola $xy=c^{2}$ is, $x+pqy=c(p+q)$

Chord of Contact

The equation of the chord of contact from the external point $(x_{0} , y_{0})$ on the rectangular hyperbola $xy=c^{2}$ is given by, $xy_{0}+yx_{0}=2c^{2}$

Geometrical Properties of the Rectangular Hyperbola

We shall now examine some examination style questions, showing the questions that may be encountered with this topic. Note that there are again two particular properties required by the syllabus, and we shall answer these as example questions.

Example 11

Prove that the equation of the chord joining the points $P(cp ,\frac{c}{p})$ and $Q(cq ,\frac{c}{q})$ on the curve $xy=c^{2}$ is $x+pqy=c(p+q)$ . If this chord is normal at $P$ , prove that $1+p^{3}q=0$ .

Solution 11

Firstly, to find the equation of the chord we must find the gradient of $PQ$ .

$m(PQ)=\frac{(\frac{c}{p}-\frac{c}{q})}{(cp-cq)}=-\frac{1}{pq}$

Now, using the point gradient formula gives,

$y-\frac{c}{p}=-\frac{1}{pq}(x-cp)$

=> $pqy-cq=-x+cp$

=> $x+pqy=c(p+q)$

Now, if the chord is normal at $P$ , then it follows that the gradient of the normal at $P$ is equal to the gradient of the chord. So, we must now find the gradient of the normal.

$y=\frac{c^{2}}{x}$

=> $\frac{dy}{dx}=-\frac{c^{2}}{x^{2}}$

At $P(cp , c/p)$ we have that,

$\frac{dy}{dx}=-\frac{c^{2}}{c^{2}p^{2}}=-\frac{1}{p^{2}}$

That is,

$m(tangent)=-\frac{1}{p^{2}}$

Now, we find the gradient of the normal by considering the negative reciprocal.

$m(normal)=-\frac{1}{(-\frac{1}{p^{2}})}=p^{2}$

$y-\frac{c}{p}=-\frac{1}{pq}(x-cp)$

=> $pqy-cq=-x+cp$

So we have that the gradient of the normal at $P$ is $p^{2}$ .

Equating this value to the gradient of $PQ$ gives,

$-\frac{1}{pq}=p^{2}p^{3}q=-1p^{3}q+1=0$

Which is the required result.

Example 12

(Area of Triangle Property) Prove that the area of the triangle bounded by the tangent and the asymptotes is a constant.

Solution 12

Consider the rectangular $xy=c^{2}$ and the point $P(cp , c/p)$ which lies upon it. Also consider the diagram below.

Since the asymptotes are $x=0$ and $y=0$ then it follows that we are to consider the area of triangle $AOB$ where $A$ and $B$ are the intercepts with the $x$ and $y$ axis respectively.Now, the tangent at the point $P(cp ,c/p)$ has equation

$x+p^{2}y=2cp$

At $y=0$ we have that

$x=2cp$

=> $A(2cp , 0)$

At $x=0$ we have that

$p^{2}y=2cp$

=> $B(0 ,\frac{2c}{p})$

Now, the area of triangle $AOB$ is given by

$Area=\frac{1}{2}(OA)(OB)$

$Area=\frac{1}{2}2cp\times \frac{2c}{p}=2c^{2}$

Hence, as $c$ is constant for any particular hyperbola, it then follows that the area of the triangle bounded by the tangent and the asymptotes is constant.

Example 13

(Length of Intercept Property) Prove that the length of the intercept cut off by the tangent and the asymptotes is twice the distance of the point of contact from the intersection of the asymptotes.

Solution 13

Consider the point $P(cp , c/p)$ and the hyperbola $xy=c^{2}$ which contains the point $P$ . Also, consider the below diagram.

We are required to show that the length of $AB$ is equal to the twice the length of $OP$ .

So, from the previous example we have the coordinates of $A$ and $B$ being, $A(2cp , 0)$ and $B(0 , 2c/p)$ . Hence we have that,

$AB=\sqrt{(2cp)^{2}+(\frac{2c}{p})^{2}}$ \\

$AB=2c\sqrt{p^{2}+\frac{1}{p^{2}}}$ Now, we also have that

$OP=\sqrt{(cp)^{2}+(\frac{c}{p})^{2}}$

$OP=c\sqrt{p^{2}+\frac{1}{p^{2}}}$

Hence we have that $AB=2 OP$ , by a comparison of the expressions they are each equivalent to. That is, the length of the intercept cut off by the tangent and the asymptotes is twice the distance of the point of contact from the intersection of the asymptotes.

Locus Problems involving the Rectangular Hyperbola

These are problems in which the locus (algebraic description of the geometrical path) of a point is required given a certain condition. In these problems, one typically eliminates the parameters using an algebraic method of elimination. It is important to note that some questions involve restrictions on the locus of the point. Example 15 is a fine show of this. We shall consider some examples to illustrate the techniques involved.

Example 14

Consider the hyperbola $xy=c^{2}$ and the points $P(cp , c/p)$ and $Q(cq , c/q)$ that lie upon this hyperbola. Find the equation of the locus of $M$ , the midpoint of the points $P$ and $Q$ given that $pq=c$ .

Solution 14

We firstly shall find the coordinates of the midpoint $M$ of $PQ$ .

$M(x , y)=(\frac{cp+cq}{2} , \frac{c}{2}(\frac{1}{p}+\frac{1}{q}))$

That is,

$M(x , y)=(\frac{c}{2}(p+q) ,\frac{c(p+q)}{2pq})$

Equating cartesian and parametric coordinates gives,

$x=\frac{c}{2}(p+q) , y=\frac{c}{2pq}(p+q)$

Now, we are given that $pq=c$ . Hence we obtain,

$x=\frac{c}{2}(p+q) , y=\frac{1}{2}(p+q)$

Now, substituting the expression for $y$ into the expression for $x$ to eliminate the parameters $p$ and $q$ gives,

$x=cy y=\frac{1}{c}x$

Hence the equation of the locus of the midpoint is the line $y=$ $\frac{1}{c}$ $x$ .

Example 15

The points $P(cp , c/p)$ and $Q(cq , c/q)$ which are in the first quadrant, lie on the hyperbola $xy=c^{2}$ . Find the locus of the intersection of the tangents at $P$ and $Q$ , given that the chord $PQ$ when extended, passes through the point $A(0 , k)$ , where $k$ is a positive real constant.

Solution 15

Firstly we draw a diagram,

The tangent at $P$ as equation

$x+p^{2}y=2cp (1)$

And the tangent at $Q$ has equation

$x+q^{2}y=2cq (2)$

$(1)-(2)$

gives,

$y(p^{2}-q^{2})=2c(p-q)y=\frac{2c(p-q)}{(p-q)(p+q)}=\frac{2c}{p+q}$

To find the $x$ value we simply make the substitution for $y$ into $(1)$ ,

$x=2cp-p^{2}(\frac{2c}{p+q})=\frac{2cp(p+q)-2cp^{2}}{p+q}=\frac{2cpq}{p+q}$

Hence we have that the point of intersection $T$ is given by the coordinates,

$T(\frac{2cpq}{p+q} ,\frac{2c}{p+q})$

Now, we find the equation of the chord $PQ$ so as to be able to use the other information regarding the point $A$ .

$m(PQ)=\frac{\frac{c}{p}-\frac{c}{q}}{cp-cq}=\frac{q-p}{(p-q)pq}=-\frac{1}{pq}$

Hence, using the point-gradient formula we obtain,

$y-\frac{c}{p}=-\frac{1}{pq}(x-cp)$

$pqy-cq=-x+cp$

=> $x+pqy=c(p+q)$

Which is the equation of the chord $PQ$ .

Now, we are given that $PQ$ , when extended passes through the point $A(0 , k)$ . Substituting into the equation for $PQ$ gives,

$0+pq(k)=c(p+q)$

=> $pq=\frac{c(p+q)}{k}$

Now, equating the coordinates of $T$ gives,

$x=\frac{2cpq}{p+q} , y=\frac{2c}{p+q}$

$0+pq(k)=c(p+q)[latex] Now, substituting the expression for [latex]pq$ into the expression for $x$ gives,

$x=\frac{(\frac{2c^{2}(p+q)}{k})}{(p+q)}=\frac{2c^{2}}{k}$

Hence as we have that the value of $x$ is a constant then it follows that the locus of the point of intersection of the tangents has equation,

$x=\frac{2c^{2}}{k}$

Now, since we have that $P$ and $Q$ lie on the first quadrant, then from the diagram it is quite obvious that the point of intersection of the tangents will at all times be within the first quadrant, and definitely below the curve $xy=c^{2}$ . Hence we have that the locus of the point $T$ is only valid for the the $y$ -values lying between $0<y\le k/2$ since we have that

$y=\frac{c^{2}}{x}=\frac{c^{2}}{(\frac{2c^{2}}{k})}=\frac{k}{2}$

$x=2c^{2}/k$

Hence the locus of the point of intersection is the line $x=2c^{2}/k$ between the $y$ -values $0<y\le k/2$ .

M	T	W	T	F	S	S
1	2	3	4	5	6	7
8	9	10	11	12	13	14
15	16	17	18	19	20	21
22	23	24	25	26	27	28
29	30

Mathematics 2B study Material( most important questions for 2019-2020 )

Telangana Intermediate 2 Year Maths Syllabus

Telangana Class 12 Maths Syllabus

Telangana Class 12 Mathematics II A Syllabus

Telangana Class 12 Mathematics II B Syllabus

Conics – Introduction

Cartesian Equations of the ellipse and hyperbola

Example 1

Solution 1

Example 2

Solution 2

Parametric Equations of the Ellipse and Hyperbola

Example 3

Solution 3

Example 4

Solution 4

Example 5

Solution 5

Example 6

Solution 6

The Tangent, Normal and Chord of Contact to the Ellipse and Hyperbola

Tangent

Ellipse

Hyperbola

Normal

Ellipse

Hyperbola

Chord of Contact

Geometrical Properties of the Hyperbola and Ellipse

Example 7

Example 8

Solution 8

Example 9

Solution 9

Example 10

Solution 10

The Rectangular Hyperbola

Proof

Parameterisation of the Rectangular Hyperbola

The Tangent, Normal, Chord and Chord of Contact to the Rectangular Hyperbola

Tangent

Normal

Chord

Chord of Contact

Geometrical Properties of the Rectangular Hyperbola

Example 11

Solution 11

Example 12

Solution 12

Example 13

Solution 13

Locus Problems involving the Rectangular Hyperbola

Example 14

Solution 14

Example 15

Solution 15

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