Joint Entrance Examination

Graduate Aptitude Test in Engineering

Strength of Materials Or Solid Mechanics

Structural Analysis

Construction Material and Management

Reinforced Cement Concrete

Steel Structures

Geotechnical Engineering

Fluid Mechanics and Hydraulic Machines

Hydrology

Irrigation

Geomatics Engineering Or Surveying

Environmental Engineering

Transportation Engineering

Engineering Mathematics

General Aptitude

1

If the parabolas y^{2} = 4b(x – c) and y^{2} = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c) ?

A

(1, 1, 3)

B

(1, 1, 0)

C

$$\left( {{1 \over 2},2,0} \right)$$

D

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

Normal to the two given curves are

y = m(x – c) – 2bm – bm^{3},

y = mx – 4am – 2am^{3}

If they have a common normal, then

(c + 2b)m + bm^{3}
= 4am + 2am^{3}

$$ \Rightarrow $$ (4a – c – 2b) m = (b – 2a)m^{3}

$$ \Rightarrow $$ (4a – c – 2b) = (b – 2a)m^{2}

$$ \Rightarrow $$ m^{2} = $${c \over {2a - b}} - 2$$ $$>$$ 0

By checking all options we found (A) is right option.

**Note :** We get that all the options are correct for m = 0
i.e., when common normal is x-axis. But may be in question they want common
normal other than x – axis, hence answer is (A).

y = m(x – c) – 2bm – bm

y = mx – 4am – 2am

If they have a common normal, then

(c + 2b)m + bm

$$ \Rightarrow $$ (4a – c – 2b) m = (b – 2a)m

$$ \Rightarrow $$ (4a – c – 2b) = (b – 2a)m

$$ \Rightarrow $$ m

By checking all options we found (A) is right option.

2

The equation of a tangent to the hyperbola 4x^{2} – 5y^{2} = 20 parallel to the line x – y = 2 is -

A

x $$-$$ y + 9 = 0

B

x $$-$$ y $$-$$ 3 = 0

C

x $$-$$ y + 1 = 0

D

x $$-$$ y + 7 = 0

Hyperbola $${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$$

slope of tangent = 1

equation of tangent y = x $$ \pm $$ $$\sqrt {5 - 4} $$

$$ \Rightarrow $$ y = x $$ \pm $$ 1

$$ \Rightarrow $$ y = x + 1

or

$$ \Rightarrow $$ y = x $$-$$ 1

slope of tangent = 1

equation of tangent y = x $$ \pm $$ $$\sqrt {5 - 4} $$

$$ \Rightarrow $$ y = x $$ \pm $$ 1

$$ \Rightarrow $$ y = x + 1

or

$$ \Rightarrow $$ y = x $$-$$ 1

3

The length of the chord of the parabola x^{2} $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$ is -

A

$$8\sqrt 2 $$

B

$$6\sqrt 3 $$

C

$$3\sqrt 2 $$

D

$$2\sqrt {11} $$

x^{2} = 4y

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

Solving together we get

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

$$\sqrt 2 $$x^{2} + 4x + 16$$\sqrt 2 $$

$$\sqrt 2 $$x^{2} $$-$$ 4x $$-$$ 16$$\sqrt 2 $$ = 0

x_{1} + x_{2} = 2$$\sqrt 2 $$; x_{1}x_{2} = $${{ - 16\sqrt 2 } \over {\sqrt 2 }}$$ = $$-$$ 16

Similarly,

($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)^{2} = 4y

2y^{2} + 32 $$-$$ 16y = 4y

$$\ell $$_{AB} = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$

$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$

$$ = \sqrt {8 + 64 + 100 - 64} $$

$$ = \sqrt {108} = 6\sqrt 3 $$

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

Solving together we get

x

$$\sqrt 2 $$x

$$\sqrt 2 $$x

x

Similarly,

($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)

2y

$$\ell $$

$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$

$$ = \sqrt {8 + 64 + 100 - 64} $$

$$ = \sqrt {108} = 6\sqrt 3 $$

4

Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents

A

an ellipse whose eccentricity is $${1 \over {\sqrt {r + 1} }},$$ where r > 1

B

an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where 0 < r < 1

C

an ellipse whose eccentricity is $${2 \over {\sqrt {r - 1} }},$$ where 0 < r < 1

D

an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where r > 1

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

for r > 1, $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$

$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$

$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$

$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$

for r > 1, $${{{y^2}} \over {1 + r}} + {{{x^2}} \over {1 - r}} = 1$$

$$e = \sqrt {1 - \left( {{{r - 1} \over {r + 2}}} \right)} $$

$$ = \sqrt {{{\left( {r + 1} \right) - \left( {r - 1} \right)} \over {\left( {r + 1} \right)}}} $$

$$ = \sqrt {{2 \over {r + 1}}} = \sqrt {{2 \over {r + 1}}} $$

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*