Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is :

A

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

B

$${\sqrt 3 }$$

C

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

D

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

$${{2{b^2}} \over a} = 8$$ and $$2b = {1 \over 2}\left( {2ae} \right)$$

$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$

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

$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$

$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$

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

$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$

2

MCQ (Single Correct Answer)

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$, is

A

$${{27 \over 2}}$$

B

$$27$$

C

$${{27 \over 4}}$$

D

$$18$$

The end point of latus rectum of ellipse

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ in first quadrant is

$$\left( {ae,{{{b^2}} \over a}} \right)$$ and the tangent at this point

intersects $$x$$-axis at $$\left( {{a \over e},0} \right)$$ and

$$y$$-axis at $$(0,a).$$

The given ellipse is $${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$$

Then $${a^2} = 9,{b^2} = 5$$

$$ \Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}$$

$$\therefore$$ end point of latus rectum in first quadrant is

$$L\left( {2,\,\,5/3} \right)$$

Equation of tangent at $$L$$ is $${{2x} \over 9} + {y \over 3} - 1$$

It meets $$x$$-axis at $$A(9/2, 0)$$

and $$y$$-axis at $$B(0,3)$$

$$\therefore$$ Area of $$\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$$

By symmetry area of quadrilateral

$$ = 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,$$ sq. units.

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ in first quadrant is

$$\left( {ae,{{{b^2}} \over a}} \right)$$ and the tangent at this point

intersects $$x$$-axis at $$\left( {{a \over e},0} \right)$$ and

$$y$$-axis at $$(0,a).$$

The given ellipse is $${{x{}^2} \over 9} + {{{y^2}} \over 5} = 1$$

Then $${a^2} = 9,{b^2} = 5$$

$$ \Rightarrow e = \sqrt {1 - {5 \over 9}} = {2 \over 3}$$

$$\therefore$$ end point of latus rectum in first quadrant is

$$L\left( {2,\,\,5/3} \right)$$

Equation of tangent at $$L$$ is $${{2x} \over 9} + {y \over 3} - 1$$

It meets $$x$$-axis at $$A(9/2, 0)$$

and $$y$$-axis at $$B(0,3)$$

$$\therefore$$ Area of $$\Delta OAB = {1 \over 2} \times {9 \over 2} \times 3 = {{27} \over 4}$$

By symmetry area of quadrilateral

$$ = 4 \times \left( {Area\,\,\Delta OAB} \right) = 4 \times {{27} \over 4} = 27\,\,$$ sq. units.

3

MCQ (Single Correct Answer)

Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :

A

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

B

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

C

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

D

$${y^2} = x$$

Let the coordinates of Q and P be (x_{1}, y_{1}) and (h, k) respectively.

$$\because$$ Q lies on x^{2} = 8y,

$$\therefore$$ x$$_1^2$$ = 8y ....... (1)

Again, P divides OQ internally in the ratio 1 : 3.

$$\therefore$$ $$h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$$ or x_{1} = 4h and

$$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$$ or y_{1} = 4k

Now putting x_{1} and y_{1} in (1) we get,

16h^{2} = 32k or, h^{2} = 2k

$$\therefore$$ the locus of P is given by, x^{2} = 2y.

4

MCQ (Single Correct Answer)

The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$

A

meets the curve again in the third quadrant.

B

meets the curve again in the fourth quadrant.

C

does not meet the curve again.

D

meets the curve again in the second quadrant.

Given curve is

$${x^2} + 2xy - 3{y^2} = 0$$

Difference $$w.r.t.x,$$

$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$

$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$

Equation of normal at $$(1,1)$$ is

$$y=2-x$$

Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$

Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$

Normal cuts the curve again in $$4$$th quadrant.

$${x^2} + 2xy - 3{y^2} = 0$$

Difference $$w.r.t.x,$$

$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$

$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$

Equation of normal at $$(1,1)$$ is

$$y=2-x$$

Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$

Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$

Normal cuts the curve again in $$4$$th quadrant.

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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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Differentiation

Application of Derivatives

Indefinite Integrals

Definite Integrals and Applications of Integrals

Differential Equations