Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

For the Hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ , which of the following remains constant when $$\alpha $$ varies$$=$$?

A

abscissae of vertices

B

abscissaeof foci

C

eccentricity

D

directrix.

Given, equation of hyperbola is $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$

We know that the equation of hyperbola is

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$

We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$

$$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $$

$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$

Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$

Hence, abscissae of foci remain constant when $$\alpha $$ varies.

We know that the equation of hyperbola is

$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$

We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$

$$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$

$$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $$

$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$

Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$

Hence, abscissae of foci remain constant when $$\alpha $$ varies.

2

MCQ (Single Correct Answer)

Angle between the tangents to the curve $$y = {x^2} - 5x + 6$$ at the points $$(2,0)$$ and $$(3,0)$$ is

A

$$\pi $$

B

$${\pi \over 2}$$

C

$${\pi \over 6}$$

D

$${\pi \over 4}$$

$${{dy} \over {dx}} = 2x - 5$$

$$\therefore$$ $${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$$

$${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$$

i.e. the tangents are perpendicular to each other.

$$\therefore$$ $${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$$

$${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$$

i.e. the tangents are perpendicular to each other.

3

MCQ (Single Correct Answer)

In the ellipse, the distance between its foci is $$6$$ and minor axis is $$8$$. Then its eccentricity is

A

$${3 \over 5}$$

B

$${1 \over 2}$$

C

$${4 \over 5}$$

D

$${1 \over {\sqrt 5 }}$$

$$2ae = 6 \Rightarrow ae = 3;\,\,2b = 8 \Rightarrow b = 4$$

$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$

$$ \Rightarrow a{}^2 = 16 + 9 = 25$$

$$ \Rightarrow a = 5$$

$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$

$${b^2} = {a^2}\left( {1 - {e^2}} \right);16 = {a^2} - {a^2}{e^2}$$

$$ \Rightarrow a{}^2 = 16 + 9 = 25$$

$$ \Rightarrow a = 5$$

$$\therefore$$ $$e = {3 \over a} = {3 \over 5}$$

4

MCQ (Single Correct Answer)

The locus of the vertices of the family of parabolas

$$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$$ is

$$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$$ is

A

$$xy = {{105} \over {64}}$$

B

$$xy = {{3} \over {4}}$$

C

$$xy = {{35} \over {16}}$$

D

$$xy = {{64} \over {105}}$$

Given parabola is $$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$$

$$ \Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a$$

$$ \Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}$$

$$\therefore$$ Vertex of parabola is $$\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)$$

To find locus of this vertex,

$$x = {{ - 3} \over {4a}}\,\,$$ and $$\,\,y = {{ - 35a} \over {16}}$$

$$ \Rightarrow a = {{ - 3} \over {4x}}\,\,$$ and $$a = - {{16y} \over {35}}$$

$$ \Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105$$

$$ \Rightarrow xy = {{105} \over {64}}$$ which is the required locus.

$$ \Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a$$

$$ \Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}$$

$$\therefore$$ Vertex of parabola is $$\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)$$

To find locus of this vertex,

$$x = {{ - 3} \over {4a}}\,\,$$ and $$\,\,y = {{ - 35a} \over {16}}$$

$$ \Rightarrow a = {{ - 3} \over {4x}}\,\,$$ and $$a = - {{16y} \over {35}}$$

$$ \Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105$$

$$ \Rightarrow xy = {{105} \over {64}}$$ which is the required locus.

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