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

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.

2

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, $$\infty $$)

3

Equation of a common tangent to the circle, x^{2} + y^{2} – 6x = 0 and the parabola, y^{2} = 4x is :

A

$$2\sqrt 3 $$y = 12x + 1

B

$$\sqrt 3 $$y = x + 3

C

$$2\sqrt 3 $$y = -x - 12

D

$$\sqrt 3 $$y = 3x + 1

We know,

Equation of tangent to the parabola y^{2} = 4ax is,

y = mx + $${a \over m}$$

$$ \therefore $$ Equation of tangent to the parabola y^{2} = 4x is,

y = mx + $${1 \over m}$$

$$ \Rightarrow $$ m^{2}x $$-$$ ym + 1 = 0

This tangent is also the tangent to the circle x^{2} + y^{2} $$-$$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$ \therefore $$ $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$

$$ \Rightarrow $$ (3m^{2} + 1)^{2} = 9(m^{4} + m^{2})

$$ \Rightarrow $$ 9m^{4} + 6m^{2} + 1 = 9m^{4} + 9m^{2}

$$ \Rightarrow $$ 3m^{2} = 1

$$ \Rightarrow $$ m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$

So, possible tangents are

y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 $$y = x + 3

or y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3

Equation of tangent to the parabola y

y = mx + $${a \over m}$$

$$ \therefore $$ Equation of tangent to the parabola y

y = mx + $${1 \over m}$$

$$ \Rightarrow $$ m

This tangent is also the tangent to the circle x

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$$ \therefore $$ $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$

$$ \Rightarrow $$ (3m

$$ \Rightarrow $$ 9m

$$ \Rightarrow $$ 3m

$$ \Rightarrow $$ m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$

So, possible tangents are

y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 $$y = x + 3

or y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$

$$ \Rightarrow $$ $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3

4

If $$\theta $$ denotes the acute angle between the curves, y = 10 – x^{2} and y = 2 + x^{2} at a point of their intersection, the |tan $$\theta $$| is equal to :

A

$$8 \over 15$$

B

$$4 \over 9$$

C

$$7 \over 17$$

D

$$8 \over 17$$

Angle between the curves is the acute angle between the tangents at the point of intersection.

y = 10 $$-$$ x

and y = 2 + x

$$ \therefore $$ 10 $$-$$ x

$$ \Rightarrow $$ 2x

$$ \Rightarrow $$ x

$$ \Rightarrow $$ x = 2, $$-$$ 2

$$ \therefore $$ points of intersection (2, 6) and ($$-$$ 2, 6)

$${{dy} \over {dx}}$$ for curve 1 = $$-$$ 2x

$$ \therefore $$ Slope(m

$${{dy} \over {dx}}$$ for curve 2 = 2x

$$ \therefore $$ slope (m

$$ \therefore $$ tan$$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$

= $$\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$$

= $${8 \over {15}}$$

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Straight Lines and Pair of Straight Lines *keyboard_arrow_right*

Circle *keyboard_arrow_right*

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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*

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