Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is

A

$${{8 \over 3}}$$

B

$${{2 \over 3}}$$

C

$${{4 \over 3}}$$

D

$${{5 \over 3}}$$

Perpendicular distance of directrix from focus

$$ = {a \over e} - ae = 4$$

$$ \Rightarrow a\left( {2 - {1 \over 2}} \right) = 4$$

$$ \Rightarrow a = {8 \over 3}$$

$$\therefore$$ Semi major axis $$=8/3$$

2

MCQ (Single Correct Answer)

The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a

A

circle

B

hyperbola

C

ellipse

D

parabola

Equation of normal at $$P\left( {x,y} \right)$$ is $$Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$$

Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)

$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$

$$ \Rightarrow X = x + y{{dy} \over {dx}}$$ $$\therefore$$ Co-ordinate of $$G\left( {x + y{{dy} \over {dx}},0} \right)$$

Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$

as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$

$$\therefore$$ $$x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$$

On Integrating $$ \Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$$

$$\therefore$$ the curve is a hyperbola

Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)

$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$

$$ \Rightarrow X = x + y{{dy} \over {dx}}$$ $$\therefore$$ Co-ordinate of $$G\left( {x + y{{dy} \over {dx}},0} \right)$$

Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$

as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$

$$\therefore$$ $$x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$$

On Integrating $$ \Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$$

$$\therefore$$ the curve is a hyperbola

3

MCQ (Single Correct Answer)

The equation of a tangent to the parabola $${y^2} = 8x$$ is $$y=x+2$$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

A

$$(2,4)$$

B

$$(-2,0)$$

C

$$(-1,1)$$

D

$$(0,2)$$

Parabola $${y^2} = 8x$$

We know that the locus of point of intersection of two perpendicular tangents to a

parabola is its directrix. Point must be on the directrix of parabola

as equation of directrix $$x + 2 = 0 \Rightarrow x = - 2$$

Hence the point is $$\left( { - 2,0} \right)$$

We know that the locus of point of intersection of two perpendicular tangents to a

parabola is its directrix. Point must be on the directrix of parabola

as equation of directrix $$x + 2 = 0 \Rightarrow x = - 2$$

Hence the point is $$\left( { - 2,0} \right)$$

4

MCQ (Single Correct Answer)

For the Hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ , which of the following remains constant when $$\alpha $$ varies$$=$$?

A

abscissae of vertices

B

abscissaeof foci

C

eccentricity

D

directrix.

Given, equation of hyperbola is $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$

We know that the equation of hyperbola is

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$

We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$

$$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $$

$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$

Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$

Hence, abscissae of foci remain constant when $$\alpha $$ varies.

We know that the equation of hyperbola is

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$

We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$

$$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $$

$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$

Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$

Hence, abscissae of foci remain constant when $$\alpha $$ varies.

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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

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Properties of Triangle

Inverse Trigonometric Functions

Straight Lines and Pair of Straight Lines

Circle

Conic Sections

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Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations