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

If y = mx + c is the normal at a point on the parabola y^{2} = 8x whose focal distance is 8 units, then $$\left| c \right|$$ is equal to :

A

$$2\sqrt 3 $$

B

$$8\sqrt 3 $$

C

$$10\sqrt 3 $$

D

$$16\sqrt 3 $$

c = $$-$$ 29m $$-$$ 9m^{3}

a = 2

Given (at^{2} $$-$$ a)^{2} + 4a^{2}t^{2} = 64

$$ \Rightarrow $$ (a(t^{2} + 1)) = 8

$$ \Rightarrow $$ t^{2} + 1 = 4 $$ \Rightarrow $$ t^{2} = 3

$$ \Rightarrow $$ t = $$\sqrt 3 $$

$$ \therefore $$ c = 2at(2 + t^{2})

= $$2\sqrt 3 \left( 5 \right)$$

$$\left| c \right|$$ = 10$$\sqrt 3 $$

a = 2

Given (at

$$ \Rightarrow $$ (a(t

$$ \Rightarrow $$ t

$$ \Rightarrow $$ t = $$\sqrt 3 $$

$$ \therefore $$ c = 2at(2 + t

= $$2\sqrt 3 \left( 5 \right)$$

$$\left| c \right|$$ = 10$$\sqrt 3 $$

2

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate
axes and passing through the points (4, −1) and (−2, 2) is :

A

$${1 \over 2}$$

B

$${2 \over {\sqrt 5 }}$$

C

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

D

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

Centre at (0, 0)

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1

at point (4, $$-$$ 1)

$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1

$$ \Rightarrow $$ 16b^{2} + a^{2} = a^{2}b^{2} . . . .(i)

at point ($$-$$ 2, 2)

$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ 4b^{2} + 4a^{2} = a^{2}b^{2} . . . .(ii)

$$ \Rightarrow $$ 16b^{2} + a^{2} = 4a^{2} + 4b^{2}

From equations (i) and (ii)

$$ \Rightarrow $$ 3a^{2} = 12b^{2}

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

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

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

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

$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}}$$ = 1

at point (4, $$-$$ 1)

$${{16} \over {{a^2}}} + {1 \over {{b^2}}}$$ = 1

$$ \Rightarrow $$ 16b

at point ($$-$$ 2, 2)

$${4 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$

$$ \Rightarrow $$ 4b

$$ \Rightarrow $$ 16b

From equations (i) and (ii)

$$ \Rightarrow $$ 3a

$$ \Rightarrow $$

b

$$ \Rightarrow $$ e

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

3

Tangents are drawn to the hyperbola 4x^{2} - y^{2} = 36 at the points P and Q.

If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is :

If these tangents intersect at the point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is :

A

$$36\sqrt 5 $$

B

$$45\sqrt 5 $$

C

$$54\sqrt 3 $$

D

$$60\sqrt 3 $$

Here PQ is the chord of contact.

Equation of PQ is

x.0 $$-$$ y.3 = 36

$$ \Rightarrow \,\,\,\,y = - 12$$

Putting value of y = $$-$$ 12 in the equation

4x

we get , 4x

$$ \Rightarrow \,\,\,\,4{x^2}\, = \,180$$

$$ \Rightarrow \,\,\,\,\,{x^2}\,\, = \,\,45$$

$$ \Rightarrow \,\,\,\,x = \pm \,3\sqrt 5 $$

$$\therefore\,\,\,$$ Coordinate of $$P = \left( { - 3\sqrt 5 , - 12} \right)$$ and

Coordinate of $$Q = \left( {3\sqrt 5 , - 12} \right)$$

$$\therefore\,\,\,$$ Length of PQ $$ = 3\sqrt 5 + 3\sqrt 5 $$

$$ = 6\sqrt 5 .$$

TM is the height of the triangle Length of TM = 12 + 3 = 15

$$\therefore\,\,\,$$ Area of $$\Delta PQT$$ = $${1 \over 2} \times 6\sqrt 5 \times 15$$

$$ = 45\sqrt 5 $$ sq. units

4

Tangent and normal are drawn at P(16, 16) on the parabola y^{2} = 16x, which intersect the axis of the
parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and $$\angle $$CPB =
$$\theta $$, then a value of tan$$\theta $$ is :

A

$${4 \over 3}$$

B

$${1 \over 2}$$

C

2

D

3

As equation of tangent PA at (x

yy

here (x

y . 16 = 2.4 (x + 16)

$$ \Rightarrow $$ 2y = x + 16 .....(1)

At pont A value of y = 0

putting y = 0 in equation (1) we get,

0 = x + 16

$$ \Rightarrow $$ x = $$-$$ 16

$$\therefore\,\,\,$$ Coordinate of point A = ($$-$$ 16, 0)

Slope of line P A :

As$$\,\,\,\,$$ 2y = x + 16

$$\therefore\,\,\,$$ y = $${1 \over 2}\,$$ x + 8

$$\therefore\,\,\,$$ Slope (m) = $${1 \over 2}\,$$

Let slope of perpendicular line PB passing through point p(16, 16) = m'

$$\therefore\,\,\,$$ m m' = $$-$$ 1

$$ \Rightarrow $$ $${1 \over 2}$$ $$ \times $$ m' = $$-$$ 1

$$ \Rightarrow $$ ' = $$-$$ 2

As Equation of normal PB, when slope is m,

y = mx $$-$$ 2am $$-$$am

Here m = m' = $$-$$ 2 and a = 4

$$\therefore\,\,\,$$ y = $$-$$2x $$-$$ 2(4) ($$-$$2) $$-$$ 4 . ($$-$$2)

$$ \Rightarrow $$ y = $$-$$ 2x + 16 + 32

$$ \Rightarrow $$ y = $$-$$ 2x + 48 ..... (2)

At point B, y = 0

puttig y = 0 at equation (2) we get,

0 = $$-$$ 2x + 48

$$ \Rightarrow $$x = 24

$$\therefore\,\,\,$$ Coordinate of point B = (24, 0)

A circle is passing through point P, A and B, and C is the center of the circle.

So, AC and BC are the radius.

Then AC = BC, So C is the middle point of line AB.

$$\therefore\,\,\,$$ C = $$\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)$$ = (4, 0)

$$\angle $$CPB = $$\theta $$ and we have to find tan $$\theta $$.

Slope of line PC = $${{16 - 0} \over {16 - 4}}$$ = $${4 \over 3}$$ =m

and we know slope of line PB = $$-$$2 = m

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

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = $$\left| { - 2} \right|$$

$$ \Rightarrow $$ tan $$\theta $$ = 2

Number in Brackets after Paper Name Indicates No of Questions

AIEEE 2002 (1) *keyboard_arrow_right*

AIEEE 2003 (2) *keyboard_arrow_right*

AIEEE 2004 (2) *keyboard_arrow_right*

AIEEE 2005 (3) *keyboard_arrow_right*

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JEE Main 2013 (Offline) (2) *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*