Mathematics 2B study Material( most important questions for 2019-2020 )
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.
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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 be a point on the conic section, and be two points called foci. Also, let and be the two points of intersection of the perpendiculars drawn from the point to two parallel lines called the directrices, with corresponding to the directrix closer to and corresponding to the directrix closer to . The locus of the point is the conic section required, where satisifes:
where is a non-negative constant called the eccentricity. |
The value of 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;
where . This form of the ellipse has a graph as shown below.
The points marked and are the two foci of the conic section. The lines marked and 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 and ) and the smaller axis of symmetry is called the “minor axis” (the line connecting and ). 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 shown before in the focus directrix definition is found to be bounded by and . That is, we have that , for the ellipse. The formulae defining the eccentricity, foci and directrices are given below for this form of the ellipse.
Component | Formula |
Eccentricity | |
Foci | |
Directrices |
The above formulae must be memorised.
The hyperbola that is most frequently studied in this course has the Cartesian equation;
Where . This form of the hyperbola can be depicted graphically as;
As before, the points marked and are the foci of the conic section, with and representing the directrices. The hyperbola also has another special feature; namely, the existence of asymptotes. As we have that and these become the oblique asymptotes of the graph. Notice here that the directrices occur closer to the -axis than the foci do in this case. We shall now study why this is so.
The eccentricity of the hyperbola is bounded below by and unbounded above. That is the eccentricity of a hyperbola is given by . 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 and that .
Component | Formula |
Eccentricity | & |
Foci | |
Directrices | |
Asymptotes |
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 and that the parabola has an eccentricity of .
We shall now look at example questions, looking at drawing graphs of these conic sections.
Example 1Sketch the graph of showing all important features. Solution 1Here, “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 . In this case and hence . Now, to find the directrices and foci, we must find the eccentricity. Using the previously shown formulae; Upon substitution into the equation we obtain; Now, to obtain the foci we simply substitute into the formula . Hence we obtain; Now, to obtain the directrices we simply substitute into the formula . We then have; Now, we must also find the intercepts. This is easily done, and we obtain as the -intercepts and as the -intercepts. Hence we obtain the graph: |
Example 2Sketch the graph of , showing all important features. Solution 2Here, important features imply foci, directrices, asymptotes and intercepts. We firstly find the eccentricity so as to be able to find other features. Now, we have that the foci are given by; Also, finding the directrices, Now, finding the asymptotes; Also, we must find the intercepts with the respective axis, to be able to sketch the required graph. Doing so gives the -intercepts as . 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 with , and the hyperbola of the form . 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 has the parametric representation:
That is, the point lying on the ellipse has parametric representation where 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 are given by,
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 3Find the parametric equations for the ellipse . Solution 3Here we have that and . Hence we have that the parametric equations are given by, upon comparison with the formulae given. |
Example 4Find the parametric equations of the hyperbola Solution 4We have that and that . Hence we have that the parametric equations of the hyperbola are given by; upon comparison with the formulae given. |
Example 5The point lies on the ellipse . Find the eccentric angle at this point (i.e. find the value of the parameter at this point). Solution 5Firstly we find the parametric equations and then simply equate the coordinates to the parametric equations. Now, we have that and . Hence we have that the parametric equations are Now, equating these parametric equations to the and values that occur at those points, we obtain; That is, Now, the only quadrant where both cosine and sine are positive is in the first quadrant. Hence we have that the eccentric angle is; |
Example 6and lie on the hyperbola . a) If subtends a right angle at , show that . b) If subtends a right angle at , show that . Solution 6a) We must firstly consider the gradients of and after which we must consider the product of the gradients which must equal to if subtends a right angle at . Now, since subtends a right angle, then it follows that . That is the product of the gradients must be . So we have that, => b) Now, we use the same technique as used above, except that we substitute the point for the point . So we have that; and that Now, we have that as before. => => Multiplying top and bottom through by gives, Now, we have that . Using this result, and the double angle formulae for sine gives, Cancelling out factors and simplifying gives, Hence we have; => Taking the reciprocal of both sides gives, |
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 at the point is |
(Parametric) The equation of the tangent at the point on the ellipse is, |
Hyperbola
(Cartesian) The equation of the tangent on the hyperbola at the point is |
(Parametric) The equation of the tangent on the hyperbola at the point is given by, |
Normal
Ellipse
(Cartesian) The equation of the normal at the point on the ellipse is, |
(Parametric) The equation of the normal at the point on the ellipse is, |
Hyperbola
(Cartesian) The equation of the normal at the point on the hyperbola is, |
(Parametric) The equation of the normal at the point on the hyperbola is, |
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 and . 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 to a curve, is simply the line connecting the two points of contact of the two tangents drawn from the point 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 is the line .
The equation of the chord of contact from the external point on the ellipse is given by, |
The equation of the chord of contact from the external point on the hyperbola is, |
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 7The hyperbola has equation . a) Find the coordinates of the foci, . b) Find the equations of the directrices. c) Find the equations of the asymptotes. d) Sketch the curve indicating the information obtained in (a) (c). e) The point lies on . Prove that the equation of the tangent at is . f) The tangent at meets the asymptotes at and . Prove that is the midpoint of . Solution 7: a) To find the coordinates of the foci, we must initially find the eccentricity, after expressing the equation in standard form. Now, using the equation for the hyperbola gives, Hence we have the foci as, b) The equations of the directrices are given by, c) The equations of the asymptotes are given by, 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. Upon implicitly differentiating with respect to , At we have the gradient is equal to, Now, at the point we have that the equation of the tangent is, Upon multiplying through by , => That is, => Now, lies on the hyperbola and hence we have that; Hence we have, Which is the equation of the tangent at the point . f) So, to find the coordinates of and we must simply solve simultaneously with the equations of the asymptotes. So consider the asymptote . => and hence we have that, Hence we have that the point is, Now, to find the point we simply solve with the equation . => => Hence we have that the point is given by, Now, consider the midpoint of . => Now, we have that lies on the hyperbola, and hence we have that => Thus we have that, Hence we have that, That is, the midpoint of is the point . |
Example 8(Reflection property) Prove that the tangent to an ellipse at a point on it, is equally inclined to the focal chords through . Solution 8Consider the ellipse and the point which lies on the ellipse. Consider the points of intersection of the tangents with the directrices being and on the directrices corresponding to the foci and respectively. Also, consider the perpendicular from to the directrices and the point of contact with the directrices being and corresponding to the foci and respectively. Also, we draw a line through parallel to the directrices of the ellipse. Now, the marked angles, namely => and =>, are the angles we are required to show are equal, since and are each focal chords.Now, from the tangent-directrix property (check previous example) the lines and subtend right angles at their respective foci, namely and respectively. This fact will later become important.Now, the two directrices and the line through 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, => So, we obtain the expression Upon rearranging we obtain Now since triangles and are each right angled, we observe that => and that Hence we have that And upon taking the inverse cosines of both sides we obtain => since the angles each lie between and degrees, as they are angles within a triangle. Hence we have that the tangent to an ellipse at a point on it, is equally inclined to the focal chords through . |
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 9We firstly draw a diagram to depict the question pictorially. Note that we are required to prove that =>. So, consider the hyperbola and the point on this hyperbola.We must initially find the coordinates of the point which is the intersection of the tangent at with the directrix corresponding to the focu . The equation of the tangent at is given by and hence when we have the point of intersection of the tangent with the directrix. Making this substitution gives, => Hence we have that the point is given by the coordinates, Now, from here we may use any of two methods to prove that =>. The first method involves use of Pythagoras’ theorem and consideration of distances. By proving that satisfies Pythagoras’theorem with as hypotenuse, we will have successfully proven that =>. Another valid method is to use the fact that the product of the gradients of and must equal if =>. 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 Now, we have that Now, recall that . Hence we have that Hence we have that =>, and hence this completes the proof. |
We shall now look at a further examination style question.
Example 10a) Show that the equation of the tangent and normal at to the ellipse are and respectively. b) The tangent and normal at cut the -axis at and respectively. Find the coordinates of and . c) Show that the focus lies on the circumference of the semi-circle which has diameter . Solution 10a) Differentiating implicitly with respect to we obtain, b) Now, substituting in the point gives, So, using the point-gradient formula to find the equation of the tangent at the point we obtain, => Hence we have that Which is the equation of the tangent at the point . 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 So, using the point-gradient formula we obtain, Upon rearranging ad dividing through by . => Thus we obtain the equation of the normal. c) We firstly draw a diagram of this situation. Firstly, for the tangent at we have the -intercept which is given by For the normal we obtain we obtain Hence we have that and are the points given by To show that the focus lies on the circumference of the semi-circle which has diameter we must show that => as this shows that is the diameter of a circle passing through by the ‘angle in a semi-circle’ rule in circle geometry. So we shall do so using gradients. Now, . So we obtain, Hence => and thus we have that lies on the circle with 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,
=>
Since . Hence we have that the rectangular hyperbola has equation of the form
where . Also, due to the fact that , we thus have that the asymptotes are .
A special property of all rectangular hyperbolae is that the eccentricity is . That is . Here is the proof;
Now, recall that for the rectangular hyperbola. Hence we have that
That is,
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 and continue on from there.
The rectangular hyperbola is in general not represented in this form, rather, it is represented in the form where . To obtain this form of the rectangular hyperbola, we translate the axes, or in other words, we rotate the curve by 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 is multiplied by the complex number , the resultant complex number is of the form . That is the resultant complex number is a rotation of the complex number on the Argand plane by an angle of . We may use this technique to great effect to show that the hyperbola converts to the hyperbola where upon rotation by .
Proof
We are performing the following transformation;
Let be the complex number to which is transformed to (consequently, the coordinates to which is transformed to are ). So we have that,
i.e.
Expanding out the RHS gives,
Now, equating real and imaginary parts gives,
Now, to eliminate and to obtain a new equation that and satisfy, we must multiply the expressions for and .
So, we have
=>
But and hence we have that
Letting we obtain the general form of the equation of the rectangular hyperbola after being rotated by .
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 | |
Directrices | |
Vertices |
The asymptotes now, are obviously and .
Parameterisation of the Rectangular Hyperbola
We shall now consider the parametric representation of the rectangular hyperbola. The rectangular hyperbola has parametric representation
where is the parameter. It can be seen that upon elimination of the parameter, the equation 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 on the rectangular hyperbola is, |
(Parametric) The equation of the tangent at the point on the rectangular hyperbola is, |
Normal
(Cartesian) The normal at the point on the rectangular hyperbola is given by, |
(Parametric) The normal at the point on the rectangular hyperbola is given by, |
Chord
The equation of the chord joining the points and on the hyperbola is, |
Chord of Contact
The equation of the chord of contact from the external point on the rectangular hyperbola is given by, |
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 11Prove that the equation of the chord joining the points and on the curve is . If this chord is normal at , prove that . Solution 11Firstly, to find the equation of the chord we must find the gradient of . Now, using the point gradient formula gives, => => Now, if the chord is normal at , then it follows that the gradient of the normal at is equal to the gradient of the chord. So, we must now find the gradient of the normal. => At we have that, That is, Now, we find the gradient of the normal by considering the negative reciprocal. => So we have that the gradient of the normal at is . Equating this value to the gradient of gives, 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 12Consider the rectangular and the point which lies upon it. Also consider the diagram below. Since the asymptotes are and then it follows that we are to consider the area of triangle where and are the intercepts with the and axis respectively.Now, the tangent at the point has equation At we have that => At we have that => Now, the area of triangle is given by Hence, as 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 13Consider the point and the hyperbola which contains the point . Also, consider the below diagram. We are required to show that the length of is equal to the twice the length of . So, from the previous example we have the coordinates of and being, and . Hence we have that, \\ Now, we also have that Hence we have that , 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 14Consider the hyperbola and the points and that lie upon this hyperbola. Find the equation of the locus of , the midpoint of the points and given that . Solution 14We firstly shall find the coordinates of the midpoint of . That is, Equating cartesian and parametric coordinates gives, Now, we are given that . Hence we obtain, Now, substituting the expression for into the expression for to eliminate the parameters and gives, Hence the equation of the locus of the midpoint is the line . |
Example 15The points and which are in the first quadrant, lie on the hyperbola . Find the locus of the intersection of the tangents at and , given that the chord when extended, passes through the point , where is a positive real constant. Solution 15Firstly we draw a diagram, The tangent at as equation And the tangent at has equation gives, To find the value we simply make the substitution for into , Hence we have that the point of intersection is given by the coordinates, Now, we find the equation of the chord so as to be able to use the other information regarding the point . Hence, using the point-gradient formula we obtain, => Which is the equation of the chord . Now, we are given that , when extended passes through the point . Substituting into the equation for gives, => Now, equating the coordinates of gives, into the expression for gives, Hence as we have that the value of is a constant then it follows that the locus of the point of intersection of the tangents has equation, Now, since we have that and 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 . Hence we have that the locus of the point is only valid for the the -values lying between since we have that at Hence the locus of the point of intersection is the line between the -values . |