Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is:

A

$$4{x^2} + 3{y^2} = 1$$

B

$$3{x^2} + 4{y^2} = 12$$

C

$$4{x^2} + 3{y^2} = 12$$

D

$$3{x^2} + 4{y^2} = 1$$

$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$

$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$

$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$

Equation of elhipe is

$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$

$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$

$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$

Equation of elhipe is

$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$

2

MCQ (Single Correct Answer)

If $$a \ne 0$$ and the line $$2bx+3cy+4d=0$$ passes through the points of intersection of the parabolas $${y^2} = 4ax$$ and $${x^2} = 4ay$$, then

A

$${d^2} + {\left( {3b - 2c} \right)^2} = 0$$

B

$${d^2} + {\left( {3b + 2c} \right)^2} = 0$$

C

$${d^2} + {\left( {2b - 3c} \right)^2} = 0$$

D

$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$

Solving equations of parabolas

$${y^2} = 4ax$$ and $${x^2} = 4ay$$

we get $$(0,0)$$ and $$(4a, 4a)$$

Substituting in the given equation of line

$$2bx+3cy+4d=0,$$

we get $$d=0$$

and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$

$${y^2} = 4ax$$ and $${x^2} = 4ay$$

we get $$(0,0)$$ and $$(4a, 4a)$$

Substituting in the given equation of line

$$2bx+3cy+4d=0,$$

we get $$d=0$$

and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$

3

MCQ (Single Correct Answer)

The foci of yhe ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is

A

$$9$$

B

$$1$$

C

$$5$$

D

$$7$$

$${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$

$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$

$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$

$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$

$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola

$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$

$$\therefore$$ $$e = {3 \over 4}$$

Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$

$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$

$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$

$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$

$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$

$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola

$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$

$$\therefore$$ $$e = {3 \over 4}$$

Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$

$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$

4

MCQ (Single Correct Answer)

The normal at the point$$\left( {bt_1^2,2b{t_1}} \right)$$ on a parabola meets the parabola again in the point $$\left( {bt_2^2,2b{t_2}} \right)$$, then

A

$${t_2} = {t_1} + {2 \over {{t_1}}}$$

B

$${t_2} = -{t_1} - {2 \over {{t_1}}}$$

C

$${t_2} = -{t_1} + {2 \over {{t_1}}}$$

D

$${t_2} = {t_1} - {2 \over {{t_1}}}$$

Equation of the normal to a parabola $${y^2} = 4bx$$ at point

$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$

As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then

$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$

$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$

$$ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$

$$ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$

$$ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$

$$ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$

$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$

As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then

$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$

$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$

$$ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$

$$ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$

$$ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$

$$ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$

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Complex Numbers

Quadratic Equation and Inequalities

Permutations and Combinations

Mathematical Induction and Binomial Theorem

Sequences and Series

Matrices and Determinants

Vector Algebra and 3D Geometry

Probability

Statistics

Mathematical Reasoning

Trigonometric Functions & Equations

Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

Functions

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations