Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geomatics Engineering Or Surveying

Engineering Mechanics

Hydrology

Transportation Engineering

Strength of Materials Or Solid Mechanics

Reinforced Cement Concrete

Steel Structures

Irrigation

Environmental Engineering

Engineering Mathematics

Structural Analysis

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

General Aptitude

1

P and Q are two distinct points on the parabola, y^{2} = 4x, with parameters t and t_{1} respectively. If the normal at P passes through Q, then the minimum value of $$t_1^2$$ is :

A

2

B

4

C

6

D

8

t_{1} = $$-$$ t $$-$$ $${2 \over t}$$

$$t_1^2$$ = t^{2} + $${4 \over {{t^2}}}$$ + 4

t^{2} + $${4 \over {{t^2}}}$$ $$ \ge $$ 2$$\sqrt {{t^2}.{4 \over {{t^2}}}} = 4$$

Minimum value of $$t_1^2$$ = 8

$$t_1^2$$ = t

t

Minimum value of $$t_1^2$$ = 8

2

A hyperbola whose transverse axis is along the major axis of the conic, $${{{x^2}} \over 3} + {{{y^2}} \over 4} = 4$$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $${3 \over 2},$$ then which of the following points does **NOT** lie on it ?

A

(0, 2)

B

$$\left( {\sqrt 5 ,2\sqrt 2 } \right)$$

C

$$\left( {\sqrt {10} ,2\sqrt 3 } \right)$$

D

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

$${{{x^2}} \over {12}} + {{{y^2}} \over {16}}$$ = 1

e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$

Foci (0, 2) & (0, $$-$$ 2)

So, transverse axis of hyperbola

= 2b = 4

$$ \Rightarrow $$ b = 2 & a^{2} = 1^{2} (e^{2} $$-$$ 1)

$$ \Rightarrow $$ a^{2} = 4$$\left( {{9 \over 4} - 1} \right)$$

$$ \Rightarrow $$ a^{2} = 5

$$ \therefore $$ It's equation is $${{{x^2}} \over 5} - {{{y^2}} \over 4}$$ = $$-$$ 1

The point (5, 2$$\sqrt 3 $$) does not satisfy the above equation.

e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$

Foci (0, 2) & (0, $$-$$ 2)

So, transverse axis of hyperbola

= 2b = 4

$$ \Rightarrow $$ b = 2 & a

$$ \Rightarrow $$ a

$$ \Rightarrow $$ a

$$ \therefore $$ It's equation is $${{{x^2}} \over 5} - {{{y^2}} \over 4}$$ = $$-$$ 1

The point (5, 2$$\sqrt 3 $$) does not satisfy the above equation.

3

A hyperbola passes through the point P$$\left( {\sqrt 2 ,\sqrt 3 } \right)$$ and has foci at $$\left( { \pm 2,0} \right)$$. Then the tangent to this hyperbola at P also passes through the point

A

$$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$$

B

$$\left( {\sqrt 3 ,\sqrt 2 } \right)$$

C

$$\left( { - \sqrt 2 , - \sqrt 3 } \right)$$

D

$$\left( {3\sqrt 2 ,2\sqrt 3 } \right)$$

Equation of hyperbola is $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$

foci is (±2, 0)

$$ \Rightarrow $$ ae = 2

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

Since b^{2} = a^{2} (e^{2} – 1)

b^{2} = a^{2} e^{2} – a^{2}

$$ \therefore $$ a^{2} + b^{2} = 4 .....(1)

Also Hyperbola passes through $$\left( {\sqrt 2 ,\sqrt 3 } \right)$$

$$ \therefore $$ $${2 \over {{a^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ $${2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ (b^{2} – 3) (b^{2} + 4) = 0

$$ \therefore $$ b^{2} = 3 or b^{2} = -4

For b^{2} = 3

$$ \Rightarrow $$ a^{2} = 1

$${{{x^2}} \over 1} - {{{y^2}} \over 3} = 1$$

Equation of tangent is $${{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1$$

It satisfy point $$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$$.

foci is (±2, 0)

$$ \Rightarrow $$ ae = 2

$$ \Rightarrow $$ a

Since b

b

$$ \therefore $$ a

Also Hyperbola passes through $$\left( {\sqrt 2 ,\sqrt 3 } \right)$$

$$ \therefore $$ $${2 \over {{a^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ $${2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ (b

$$ \therefore $$ b

For b

$$ \Rightarrow $$ a

$${{{x^2}} \over 1} - {{{y^2}} \over 3} = 1$$

Equation of tangent is $${{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1$$

It satisfy point $$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$$.

4

The eccentricity of an ellipse whose centre is at the origin is $${1 \over 2}$$. If one of its directrices is x = – 4, then the
equation of the normal to it at $$\left( {1,{3 \over 2}} \right)$$ is

A

2y – x = 2

B

4x – 2y = 1

C

4x + 2y = 7

D

x + 2y = 4

Given e = $${1 \over 2}$$ and $${a \over e}$$ = 4

$$ \therefore $$ $$a$$ = 2

We have b^{2} = $$a$$^{2} (1 – e^{2}) = $$4\left( {1 - {1 \over 4}} \right)$$ = 3

$$ \therefore $$ Equation of ellipse is

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

Now, the equation of normal at $$\left( {1,{3 \over 2}} \right)$$ is

$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$

$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$

$$ \Rightarrow $$ 4x – 2y = 1

$$ \therefore $$ $$a$$ = 2

We have b

$$ \therefore $$ Equation of ellipse is

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

Now, the equation of normal at $$\left( {1,{3 \over 2}} \right)$$ is

$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}$$

$$ \Rightarrow $$ $${{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3$$

$$ \Rightarrow $$ 4x – 2y = 1

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (1) *keyboard_arrow_right*

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Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *keyboard_arrow_right*

Complex Numbers *keyboard_arrow_right*

Quadratic Equation and Inequalities *keyboard_arrow_right*

Permutations and Combinations *keyboard_arrow_right*

Mathematical Induction and Binomial Theorem *keyboard_arrow_right*

Sequences and Series *keyboard_arrow_right*

Matrices and Determinants *keyboard_arrow_right*

Vector Algebra and 3D Geometry *keyboard_arrow_right*

Probability *keyboard_arrow_right*

Statistics *keyboard_arrow_right*

Mathematical Reasoning *keyboard_arrow_right*

Functions *keyboard_arrow_right*

Limits, Continuity and Differentiability *keyboard_arrow_right*

Differentiation *keyboard_arrow_right*

Application of Derivatives *keyboard_arrow_right*

Indefinite Integrals *keyboard_arrow_right*

Definite Integrals and Applications of Integrals *keyboard_arrow_right*

Differential Equations *keyboard_arrow_right*

Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *keyboard_arrow_right*