Joint Entrance Examination

Graduate Aptitude Test in Engineering

Geotechnical Engineering

Transportation Engineering

Irrigation

Engineering Mathematics

Construction Material and Management

Fluid Mechanics and Hydraulic Machines

Hydrology

Environmental Engineering

Engineering Mechanics

Structural Analysis

Reinforced Cement Concrete

Steel Structures

Geomatics Engineering Or Surveying

General Aptitude

1

Let P be a point on the parabola, x^{2} = 4y. If the distance of P from the center of the circle, x^{2} + y^{2} + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is :

A

x + 4y $$-$$ 2 = 0

B

x $$-$$ y + 3 = 0

C

x + y +1 = 0

D

x + 2y = 0

Let P(2t, t^{2}) be any point on the parabola.

Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$

Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t

$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$

$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$

$$ \Rightarrow $$ t^{3} + 2t + 3 = 0

$$ \Rightarrow $$ (t+1) (t^{2} $$-$$ t + 3) = 0

$$\therefore\,\,\,$$ Real roots of above equation is

t = $$-$$ 1

Coordinate of P = (2t, t^{2}) = ($$-$$2, 1)

Slope of tangent to parabola at P = t = $$-$$ 1

Therefore, equation of tangent is :

(y $$-$$ 1) = ($$-$$ 1) (x + 2)

$$ \Rightarrow $$ x + y + 1 = 0

Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$

Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t

$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$

$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$

$$ \Rightarrow $$ t

$$ \Rightarrow $$ (t+1) (t

$$\therefore\,\,\,$$ Real roots of above equation is

t = $$-$$ 1

Coordinate of P = (2t, t

Slope of tangent to parabola at P = t = $$-$$ 1

Therefore, equation of tangent is :

(y $$-$$ 1) = ($$-$$ 1) (x + 2)

$$ \Rightarrow $$ x + y + 1 = 0

2

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus an its nearest vertex on the major axis is $${3 \over 2}$$ units, then its eccentricity is :

A

$${1 \over 2}$$

B

$${1 \over 3}$$

C

$${2 \over 3}$$

D

$${1 \over 9}$$

If the cordinate of focus and vertex are (ae, 0) and (a, 0) respectively,

then distance between focus and vertex,

a $$-$$ ae = $${3 \over 2}$$ (given)

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) = $${3 \over 2}$$

Length of latus rectum,

$${{2{b^2}} \over a} = 4$$

$$ \Rightarrow $$ $$\,\,\,$$ b^{2} = 2a

$$ \Rightarrow $$ $$\,\,\,$$ a^{2}(1 $$-$$ e^{2}) = 2a [As b^{2} = a^{2} (1 $$-$$ e^{2})]

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) ( 1 + e) = 2

Putting a (1 $$-$$ e) = $${3 \over 2}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${3 \over 2}$$ (1 + e) = 2

$$ \Rightarrow $$ $$\,\,\,$$ 3 + 3e = 4

$$ \Rightarrow $$ $$\,\,\,$$ e = $${1 \over 3}$$

then distance between focus and vertex,

a $$-$$ ae = $${3 \over 2}$$ (given)

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) = $${3 \over 2}$$

Length of latus rectum,

$${{2{b^2}} \over a} = 4$$

$$ \Rightarrow $$ $$\,\,\,$$ b

$$ \Rightarrow $$ $$\,\,\,$$ a

$$ \Rightarrow $$ $$\,\,\,$$ a (1 $$-$$ e) ( 1 + e) = 2

Putting a (1 $$-$$ e) = $${3 \over 2}$$

$$ \Rightarrow $$ $$\,\,\,$$ $${3 \over 2}$$ (1 + e) = 2

$$ \Rightarrow $$ $$\,\,\,$$ 3 + 3e = 4

$$ \Rightarrow $$ $$\,\,\,$$ e = $${1 \over 3}$$

3

Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the
origin, on the positive x-axis then which of the following points does not lie on it ?

A

(5, 2$$\sqrt 6$$)

B

(6, 4$$\sqrt 2$$)

C

(8, 6)

D

(4, -4)

So the equation of the parabola,

$${\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)$$

$$ \Rightarrow $$ y

$$ \Rightarrow $$ y

By checking each options you can see. point (8, 6) does not lie on the parabola.

4

Let $$0 < \theta < {\pi \over 2}$$. If the eccentricity of the

hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater

than 2, then the length of its latus rectum lies in the interval :

hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater

than 2, then the length of its latus rectum lies in the interval :

A

(3, $$\infty $$)

B

$$\left( {{3 \over 2},2} \right]$$

C

$$\left( {1,{3 \over 2}} \right]$$

D

$$\left( {2,3} \right]$$

$$\infty $$

Given hyperbola,

$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$

here a = cos$$\theta $$

and b = sin$$\theta $$

We know, eccentricity of the hyperbola is,

$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$

$$ \therefore $$ Here eccentricity

(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$

Given that,

$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$

$$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$

$$ \Rightarrow $$ 1 + tan^{2}$$\theta $$ > 4

$$ \Rightarrow $$ tan^{2}$$\theta $$ > 3

$$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$

As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$

possible value of tan$$\theta $$ > $$\sqrt 3 $$

So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$

We know latus ractum (LR) = $${{2{b^2}} \over a}$$

$$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$

= 2 tan$$\theta $$ sin$$\theta $$

We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.

So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.

$$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$

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

= 3

Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$

= 2$$\left( \infty \right) \times 1$$

= $$\infty $$

$$ \therefore $$ Internal of LR = (3, $$\infty $$)

$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$

here a = cos$$\theta $$

and b = sin$$\theta $$

We know, eccentricity of the hyperbola is,

$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$

$$ \therefore $$ Here eccentricity

(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$

Given that,

$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$

$$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$

$$ \Rightarrow $$ 1 + tan

$$ \Rightarrow $$ tan

$$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$

As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$

possible value of tan$$\theta $$ > $$\sqrt 3 $$

So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$

We know latus ractum (LR) = $${{2{b^2}} \over a}$$

$$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$

= 2 tan$$\theta $$ sin$$\theta $$

We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.

So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.

$$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$

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

= 3

Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$

= 2$$\left( \infty \right) \times 1$$

= $$\infty $$

$$ \therefore $$ Internal of LR = (3, $$\infty $$)

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

Conic Sections *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*

Trigonometric Functions & Equations *keyboard_arrow_right*

Properties of Triangle *keyboard_arrow_right*

Inverse Trigonometric Functions *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*