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

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# Telangana Intermediate 2 Year Maths Syllabus

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:

1. Mathematics II A
2. 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 $Latex formula$ be a point on the conic section, $Latex formula$ and $Latex formula$ be two points called foci. Also, let $Latex formula$ and $Latex formula$ be the two points $Latex formula$ of intersection of the perpendiculars drawn from the point to two parallel lines called the directrices, with $Latex formula$ corresponding to the directrix closer to $Latex formula$ and $Latex formula$ corresponding to the directrix closer to $Latex formula$. The locus of the point $Latex formula$ is the conic section required, where $Latex formula$ satisifes:$Latex formula$ where $Latex formula$ is a non-negative constant called the eccentricity.

The value of $Latex formula$ 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;

$Latex formula$

where $Latex formula$. This form of the ellipse has a graph as shown below.

The points marked $Latex formula$ and $Latex formula$ are the two foci of the conic section. The lines marked $Latex formula$ and $Latex formula$ 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 $Latex formula$ and $Latex formula$ ) and the smaller axis of symmetry is called the “minor axis” (the line connecting $Latex formula$ and $Latex formula$). 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 $Latex formula$ shown before in the focus directrix definition is found to be bounded by $Latex formula$ and $Latex formula$. That is, we have that $Latex formula$, for the ellipse. The formulae defining the eccentricity, foci and directrices are given below for this form of the ellipse.

 Component Formula Eccentricity $Latex formula$ $Latex formula$ Foci $Latex formula$ Directrices $Latex formula$

The above formulae must be memorised.

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

$Latex formula$

Where $Latex formula$. This form of the hyperbola can be depicted graphically as;

As before, the points marked $Latex formula$ and $Latex formula$ are the foci of the conic section, with $Latex formula$ and $Latex formula$ representing the directrices. The hyperbola also has another special feature; namely, the existence of asymptotes. As $Latex formula$ we have that $Latex formula$ and these become the oblique asymptotes of the graph. Notice here that the directrices occur closer to the $Latex formula$-axis than the foci do in this case. We shall now study why this is so.

The eccentricity $Latex formula$ of the hyperbola is bounded below by $Latex formula$ and unbounded above. That is the eccentricity of a hyperbola is given by $Latex formula$. 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 $Latex formula$ and that $Latex formula$$Latex formula$.

 Component Formula Eccentricity $Latex formula$ & $Latex formula$ Foci $Latex formula$ Directrices $Latex formula$ Asymptotes $Latex formula$

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 $Latex formula$ and that the parabola has an eccentricity of $Latex formula$.

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

### Example 1

Sketch the graph of $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ 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 $Latex formula$. In this case $Latex formula$ and hence $Latex formula$.

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

$Latex formula$

Upon substitution into the equation we obtain;

$Latex formula$

$Latex formula$

Now, to obtain the foci we simply substitute into the formula $Latex formula$. Hence we obtain;

$Latex formula$

Now, to obtain the directrices we simply substitute into the formula $Latex formula$. We then have;

$Latex formula$

Now, we must also find the intercepts. This is easily done, and we obtain $Latex formula$ as the $Latex formula$-intercepts and $Latex formula$ as the $Latex formula$-intercepts.

Hence we obtain the graph:

### Example 2

Sketch the graph of $Latex formula$, 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.

$Latex formula$

$Latex formula$

Now, we have that the foci are given by;

$Latex formula$

Also, finding the directrices,

$Latex formula$

Now, finding the asymptotes;

$Latex formula$

Also, we must find the intercepts with the respective axis, to be able to sketch the required graph. Doing so gives the $Latex formula$-intercepts as $Latex formula$.

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 $Latex formula$ with $Latex formula$, and the hyperbola of the form $Latex formula$. 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 $Latex formula$ has the parametric representation:

 $Latex formula$

That is, the point $Latex formula$ lying on the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ has parametric representation $Latex formula$ where $Latex formula$ 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 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ are given by,

 $Latex formula$

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 $Latex formula$ $Latex formula$ $Latex formula$.

### Solution 3

Here we have that $Latex formula$ and $Latex formula$. Hence we have that the parametric equations are given by,

$Latex formula$

upon comparison with the formulae given.

### Example 4

Find the parametric equations of the hyperbola $Latex formula$ $Latex formula$ $Latex formula$$Latex formula$

### Solution 4

We have that $Latex formula$ and that $Latex formula$. Hence we have that the parametric equations of the hyperbola are given by;

$Latex formula$

upon comparison with the formulae given.

### Example 5

The point $Latex formula$ lies on the ellipse $Latex formula$ $Latex formula$ $Latex formula$. 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 $Latex formula$ and $Latex formula$. Hence we have that the parametric equations are

$Latex formula$

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

$Latex formula$

That is,

$Latex formula$

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

$Latex formula$

### Example 6

$Latex formula$ and $Latex formula$ lie on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$.

a) If $Latex formula$ subtends a right angle at $Latex formula$, show that $Latex formula$ $Latex formula$ $Latex formula$.

b) If $Latex formula$ subtends a right angle at $Latex formula$, show that $Latex formula$ $Latex formula$.

### Solution 6

a) We must firstly consider the gradients of $Latex formula$ and $Latex formula$ after which we must consider the product of the gradients which must equal to $Latex formula$ if $Latex formula$ subtends a right angle at $Latex formula$.

$Latex formula$

$Latex formula$

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

$Latex formula$

=>$Latex formula$

$Latex formula$

b) Now, we use the same technique as used above, except that we substitute the point $Latex formula$ for the point $Latex formula$. So we have that;

$Latex formula$

and that

$Latex formula$

Now, we have that

$Latex formula$

as before.

=>$Latex formula$

=>$Latex formula$

Multiplying top and bottom through by $Latex formula$ gives,

$Latex formula$

Now, we have that $Latex formula$. Using this result, and the double angle formulae for sine gives,

$Latex formula$

Cancelling out factors and simplifying gives,

$Latex formula$

Hence we have;

=>$Latex formula$

Taking the reciprocal of both sides gives,

$Latex formula$

## 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 $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$at the point $Latex formula$ is$Latex formula$

 (Parametric) The equation of the tangent at the point $Latex formula$ on the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

#### Hyperbola

 (Cartesian) The equation of the tangent on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ at the point $Latex formula$ is$Latex formula$

 (Parametric) The equation of the tangent on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ at the point $Latex formula$ is given by,$Latex formula$

### Normal

#### Ellipse

 (Cartesian) The equation of the normal at the point $Latex formula$ on the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

 (Parametric) The equation of the normal at the point $Latex formula$ on the ellipse $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

#### Hyperbola

 (Cartesian) The equation of the normal at the point $Latex formula$ on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

 (Parametric) The equation of the normal at the point $Latex formula$ on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

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 $Latex formula$ and $Latex formula$. 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 $Latex formula$ to a curve, is simply the line connecting the two points of contact of the two tangents drawn from the point $Latex formula$ to the curve. Observe the diagram below.

In this diagram, the chord of contact from the external point $Latex formula$ is the line $Latex formula$.

 The equation of the chord of contact from the external point $Latex formula$ on the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ is given by,$Latex formula$

 The equation of the chord of contact from the external point $Latex formula$ on the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ is,$Latex formula$

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 $Latex formula$ has equation $Latex formula$.

a) Find the coordinates of the foci, $Latex formula$.

b) Find the equations of the directrices.

c) Find the equations of the asymptotes.

d) Sketch the curve $Latex formula$ indicating the information obtained in (a) $Latex formula$ (c).

e) The point $Latex formula$ lies on $Latex formula$. Prove that the equation of the tangent at $Latex formula$ is $Latex formula$.

f) The tangent at $Latex formula$ meets the asymptotes at $Latex formula$ and $Latex formula$. Prove that $Latex formula$ is the midpoint of $Latex formula$.

Solution 7:

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

$Latex formula$

Now, using the equation for the hyperbola gives,

$Latex formula$

$Latex formula$

Hence we have the foci as,

b) The equations of the directrices are given by,

$Latex formula$

c) The equations of the asymptotes are given by,

$Latex formula$

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.

$Latex formula$

Upon implicitly differentiating with respect to $Latex formula$,

$Latex formula$

$Latex formula$

At $Latex formula$ we have the gradient is equal to,

$Latex formula$

Now, at the point $Latex formula$ we have that the equation of the tangent is,

$Latex formula$

Upon multiplying through by $Latex formula$,

=>$Latex formula$

That is,

$Latex formula$

=>$Latex formula$

Now, $Latex formula$ lies on the hyperbola and hence we have that;

$Latex formula$

Hence we have,

$Latex formula$

Which is the equation of the tangent at the point $Latex formula$.

f) So, to find the coordinates of $Latex formula$ and $Latex formula$ we must simply solve simultaneously with the equations of the asymptotes. So consider the asymptote $Latex formula$.

$Latex formula$

=>$Latex formula$

$Latex formula$

and hence we have that,

$Latex formula$

Hence we have that the point $Latex formula$ is,

$Latex formula$

Now, to find the point $Latex formula$ we simply solve with the equation $Latex formula$.

$Latex formula$

=>$Latex formula$

=>$Latex formula$

Hence we have that the point $Latex formula$ is given by,

$Latex formula$

Now, consider the midpoint of $Latex formula$.

$Latex formula$

$Latex formula$

=>$Latex formula$

Now, we have that $Latex formula$ lies on the hyperbola, and hence we have that

$Latex formula$

=>$Latex formula$

Thus we have that,

$Latex formula$

Hence we have that,

$Latex formula$

That is, the midpoint of $Latex formula$ is the point $Latex formula$.

### Example 8

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

### Solution 8

Consider the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ and the point $Latex formula$ which lies on the ellipse. Consider the points of intersection of the tangents with the directrices being $Latex formula$ and $Latex formula$ on the directrices corresponding to the foci $Latex formula$ and $Latex formula$ respectively. Also, consider the perpendicular from $Latex formula$ to the directrices and the point of contact with the directrices being $Latex formula$ and $Latex formula$ corresponding to the foci $Latex formula$ and $Latex formula$ respectively. Also, we draw a line through $Latex formula$ parallel to the directrices of the ellipse.

Now, the marked angles, namely =>$Latex formula$ and =>$Latex formula$, are the angles we are required to show are equal, since $Latex formula$ and $Latex formula$ are each focal chords.Now, from the tangent-directrix property (check previous example) the lines $Latex formula$ and $Latex formula$ subtend right angles at their respective foci, namely $Latex formula$ and $Latex formula$ respectively. This fact will later become important.Now, the two directrices and the line through $Latex formula$ 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,

$Latex formula$

=>$Latex formula$

So, we obtain the expression

$Latex formula$

Upon rearranging we obtain

$Latex formula$

Now since triangles $Latex formula$ and $Latex formula$ are each right angled, we observe that

=>$Latex formula$

and that

$Latex formula$

Hence we have that

$Latex formula$

And upon taking the inverse cosines of both sides we obtain

=>$Latex formula$

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

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 =>$Latex formula$.

So, consider the hyperbola $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ and the point $Latex formula$ on this hyperbola.We must initially find the coordinates of the point $Latex formula$ which is the intersection of the tangent at $Latex formula$ with the directrix corresponding to the focu $Latex formula$. The equation of the tangent at $Latex formula$ is given by

$Latex formula$

and hence when $Latex formula$ $Latex formula$ we have the point of intersection of the tangent with the directrix. Making this substitution gives,

$Latex formula$

=>$Latex formula$

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

$Latex formula$

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

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

$Latex formula$

$Latex formula$

Now, we have that

$Latex formula$

$Latex formula$

Now, recall that $Latex formula$. Hence we have that

$Latex formula$

Hence we have that =>$Latex formula$, 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 $Latex formula$ to the ellipse $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ are $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ and $Latex formula$ $Latex formula$ $Latex formula$ $Latex formula$ respectively.

b) The tangent and normal at $Latex formula$ cut the $Latex formula$-axis at$Latex formula$ and $Latex formula$ respectively. Find the coordinates of $Latex formula$ and $Latex formula$.

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

### Solution 10

a) Differentiating implicitly with respect to $Latex formula$ we obtain,

b)

$Latex formula$

Now, substituting in the point $Latex formula$ gives,

$Latex formula$

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

$Latex formula$

$Latex formula$

=>$Latex formula$

Hence we have that

$Latex formula$

Which is the equation of the tangent at the point $Latex formula$. 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

$Latex formula$

So, using the point-gradient formula we obtain,

$Latex formula$

$Latex formula$

Upon rearranging ad dividing through by $Latex formula$.

=>$Latex formula$

Thus we obtain the equation of the normal.

c) We firstly draw a diagram of this situation.

Firstly, for the tangent at $Latex formula$ we have the $Latex formula$-intercept which is given by

$Latex formula$

For the normal we obtain

$Latex formula$ we obtain$Latex formula$

Hence we have that $Latex formula$ and $Latex formula$ are the points given by

$Latex formula$

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

So we shall do so using gradients.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Now, $Latex formula$. So we obtain,

$Latex formula$

Hence =>$Latex formula$ and thus we have that $Latex formula$ lies on the circle with $Latex formula$ 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,

$Latex formula$

=>$Latex formula$

Since $Latex formula$. Hence we have that the rectangular hyperbola has equation of the form

$Latex formula$

where $Latex formula$. Also, due to the fact that $Latex formula$, we thus have that the asymptotes are $Latex formula$.

A special property of all rectangular hyperbolae is that the eccentricity is $Latex formula$. That is $Latex formula$. Here is the proof;

$Latex formula$

Now, recall that $Latex formula$ for the rectangular hyperbola. Hence we have that

$Latex formula$

That is,

$Latex formula$

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 $Latex formula$ and continue on from there.

The rectangular hyperbola is in general not represented in this form, rather, it is represented in the form $Latex formula$ where $Latex formula$. To obtain this form of the rectangular hyperbola, we translate the axes, or in other words, we rotate the curve by $Latex formula$ 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 $Latex formula$ is multiplied by the complex number $Latex formula$, the resultant complex number is of the form $Latex formula$. That is the resultant complex number is a rotation of the complex number on the Argand plane by an angle of $Latex formula$. We may use this technique to great effect to show that the hyperbola $Latex formula$ converts to the hyperbola $Latex formula$ where $Latex formula$ upon rotation by $Latex formula$.

### Proof

We are performing the following transformation;

Let $Latex formula$ be the complex number to which $Latex formula$ is transformed to (consequently, the coordinates to which $Latex formula$ is transformed to are $Latex formula$ ). So we have that,

$Latex formula$

i.e.

$Latex formula$

Expanding out the RHS gives,

$Latex formula$

Now, equating real and imaginary parts gives,

$Latex formula$

Now, to eliminate $Latex formula$ and $Latex formula$ to obtain a new equation that $Latex formula$ and $Latex formula$ satisfy, we must multiply the expressions for $Latex formula$ and $Latex formula$.

So, we have

$Latex formula$

=>$Latex formula$

But $Latex formula$ and hence we have that

$Latex formula$

Letting $Latex formula$ we obtain the general form of the equation of the rectangular hyperbola after being rotated by $Latex formula$.

 $Latex formula$

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 $Latex formula$ Directrices $Latex formula$ Vertices $Latex formula$

The asymptotes now, are obviously $Latex formula$ and $Latex formula$.

## Parameterisation of the Rectangular Hyperbola

We shall now consider the parametric representation of the rectangular hyperbola. The rectangular hyperbola $Latex formula$ has parametric representation

$Latex formula$

where $Latex formula$ is the parameter. It can be seen that upon elimination of the parameter, the equation $Latex formula$ 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 $Latex formula$ on the rectangular hyperbola$Latex formula$ is,$Latex formula$

 (Parametric) The equation of the tangent at the point $Latex formula$ on the rectangular hyperbola $Latex formula$ is,$Latex formula$

### Normal

 (Cartesian) The normal at the point $Latex formula$ on the rectangular hyperbola $Latex formula$ is given by,$Latex formula$

 (Parametric) The normal at the point $Latex formula$ on the rectangular hyperbola $Latex formula$ is given by,$Latex formula$

### Chord

 The equation of the chord joining the points $Latex formula$ and $Latex formula$ on the hyperbola $Latex formula$ is,$Latex formula$

### Chord of Contact

 The equation of the chord of contact from the external point $Latex formula$ on the rectangular hyperbola $Latex formula$ is given by,$Latex formula$

## 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 $Latex formula$ and $Latex formula$ on the curve $Latex formula$ is $Latex formula$. If this chord is normal at $Latex formula$, prove that $Latex formula$.

### Solution 11

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

$Latex formula$

Now, using the point gradient formula gives,

$Latex formula$

=>$Latex formula$

=>$Latex formula$

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

$Latex formula$

=>$Latex formula$

At $Latex formula$ we have that,

$Latex formula$

That is,

$Latex formula$

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

$Latex formula$

$Latex formula$

=>$Latex formula$

So we have that the gradient of the normal at $Latex formula$ is $Latex formula$.

Equating this value to the gradient of $Latex formula$ gives,

$Latex formula$

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 $Latex formula$ and the point $Latex formula$ which lies upon it. Also consider the diagram below.

Since the asymptotes are $Latex formula$ and $Latex formula$ then it follows that we are to consider the area of triangle $Latex formula$ where $Latex formula$ and $Latex formula$ are the intercepts with the $Latex formula$ and $Latex formula$ axis respectively.Now, the tangent at the point $Latex formula$ has equation

$Latex formula$

At $Latex formula$ we have that

$Latex formula$

=>$Latex formula$

At $Latex formula$ we have that

$Latex formula$

=>$Latex formula$

Now, the area of triangle $Latex formula$ is given by

$Latex formula$

$Latex formula$

Hence, as $Latex formula$ 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 $Latex formula$ and the hyperbola $Latex formula$ which contains the point $Latex formula$. Also, consider the below diagram.

We are required to show that the length of $Latex formula$ is equal to the twice the length of $Latex formula$.

So, from the previous example we have the coordinates of $Latex formula$ and $Latex formula$ being, $Latex formula$ and $Latex formula$. Hence we have that,

$Latex formula$\\

$Latex formula$Now, we also have that

$Latex formula$

$Latex formula$

Hence we have that $Latex formula$, 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 $Latex formula$ and the points $Latex formula$ and $Latex formula$ that lie upon this hyperbola. Find the equation of the locus of $Latex formula$, the midpoint of the points $Latex formula$ and $Latex formula$ given that $Latex formula$.

### Solution 14

We firstly shall find the coordinates of the midpoint $Latex formula$ of $Latex formula$.

$Latex formula$

That is,

$Latex formula$

Equating cartesian and parametric coordinates gives,

$Latex formula$

Now, we are given that $Latex formula$. Hence we obtain,

$Latex formula$

Now, substituting the expression for $Latex formula$ into the expression for $Latex formula$ to eliminate the parameters $Latex formula$ and $Latex formula$ gives,

$Latex formula$

Hence the equation of the locus of the midpoint is the line $Latex formula$ $Latex formula$ $Latex formula$.

### Example 15

The points $Latex formula$ and $Latex formula$ which are in the first quadrant, lie on the hyperbola $Latex formula$. Find the locus of the intersection of the tangents at $Latex formula$ and $Latex formula$, given that the chord $Latex formula$ when extended, passes through the point $Latex formula$, where $Latex formula$ is a positive real constant.

### Solution 15

Firstly we draw a diagram,

The tangent at $Latex formula$ as equation

$Latex formula$

And the tangent at $Latex formula$ has equation

$Latex formula$

$Latex formula$

gives,

$Latex formula$

To find the $Latex formula$ value we simply make the substitution for $Latex formula$ into $Latex formula$,

$Latex formula$

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

$Latex formula$

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

$Latex formula$

Hence, using the point-gradient formula we obtain,

$Latex formula$

$Latex formula$

=>$Latex formula$

Which is the equation of the chord $Latex formula$.

Now, we are given that $Latex formula$, when extended passes through the point $Latex formula$. Substituting into the equation for $Latex formula$ gives,

$Latex formula$

=>$Latex formula$

Now, equating the coordinates of $Latex formula$ gives,

$Latex formula$

$Latex formula$ into the expression for $Latex formula$ gives,

$Latex formula$

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

$Latex formula$

Now, since we have that $Latex formula$ and $Latex formula$ 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 $Latex formula$. Hence we have that the locus of the point $Latex formula$ is only valid for the the $Latex formula$-values lying between $Latex formula$ since we have that

$Latex formula$

at

$Latex formula$

Hence the locus of the point of intersection is the line $Latex formula$ between the $Latex formula$-values $Latex formula$.