Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

Let $$P$$ be the point on the parabola, $${{y^2} = 8x}$$ which is at a minimum distance from the centre $$C$$ of the circle, $${x^2} + {\left( {y + 6} \right)^2} = 1$$. Then the equation of the circle, passing through $$C$$ and having its centre at $$P$$ is:

A

$${{x^2} + {y^2} - {x \over 4} + 2y - 24 = 0}$$

B

$${{x^2} + {y^2} - 4x + 9y + 18 = 0}$$

C

$${{x^2} + {y^2} - 4x + 8y + 12 = 0}$$

D

$${{x^2} + {y^2} - x + 4y - 12 = 0}$$

Minimum distance $$ \Rightarrow $$ perpendicular distance

$$E{q^n}$$ of normal at $$p\left( {2{t^2},\,4t} \right)$$

$$y = - tx + 4t + 2{t^3}$$

It passes through $$C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0$$

$$ \Rightarrow t = - 1$$

Center of new circle $$ = P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)$$

Radius $$ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 $$

$$\therefore$$ Equation of the circle is

$${\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}$$

$$ \Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0$$

$$E{q^n}$$ of normal at $$p\left( {2{t^2},\,4t} \right)$$

$$y = - tx + 4t + 2{t^3}$$

It passes through $$C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0$$

$$ \Rightarrow t = - 1$$

Center of new circle $$ = P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)$$

Radius $$ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 $$

$$\therefore$$ Equation of the circle is

$${\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}$$

$$ \Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0$$

2

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 $$2b = {1 \over 2}\left( {2ae} \right)$$

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

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

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

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

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

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

3

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.

4

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.

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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