Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

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.

2

MCQ (Single Correct Answer)

The locus of a point $$P\left( {\alpha ,\beta } \right)$$ moving under the condition that the line $$y = \alpha x + \beta $$ is tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is

A

an ellipse

B

a circle

C

a parabola

D

a hyperbola

Tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is

$$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$

Given that $$y = \alpha x + \beta $$ is the tangent of hyperbola

$$ \Rightarrow m = \alpha $$ and $${a^2}{m^2} - {b^2} = {\beta ^2}$$

$$\therefore$$ $${a^2}{\alpha ^2} - {b^2} = {\beta ^2}$$

Locus is $${a^2}{x^2} - {y^2} = {b^2}$$ which is hyperbola.

$$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$

Given that $$y = \alpha x + \beta $$ is the tangent of hyperbola

$$ \Rightarrow m = \alpha $$ and $${a^2}{m^2} - {b^2} = {\beta ^2}$$

$$\therefore$$ $${a^2}{\alpha ^2} - {b^2} = {\beta ^2}$$

Locus is $${a^2}{x^2} - {y^2} = {b^2}$$ which is hyperbola.

3

MCQ (Single Correct Answer)

Let $$P$$ be the point $$(1, 0)$$ and $$Q$$ a point on the parabola $${y^2} = 8x$$. The locus of mid point of $$PQ$$ is

A

$${y^2} - 4x + 2 = 0$$

B

$${y^2} + 4x + 2 = 0$$

C

$${x^2} + 4y + 2 = 0$$

D

$${x^2} - 4y + 2 = 0$$

$$P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)$$ Such that $${k^2} = 8h$$

Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$

$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$

$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$

$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$

$$ \Rightarrow {y^2} - 4x + 2 = 0.$$

Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$

$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$

$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$

$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$

$$ \Rightarrow {y^2} - 4x + 2 = 0.$$

4

MCQ (Single Correct Answer)

An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F$$' its focii and theangle $$FBF$$' is a right angle. Then the eccentricity of the ellipse is

A

$${1 \over {\sqrt 2 }}$$

B

$${1 \over 2}$$

C

$${1 \over 4}$$

D

$${1 \over {\sqrt 3 }}$$

as $$\angle FBF' = {90^ \circ }$$

$$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$$

$$\therefore$$ $${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$$

$$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$$

$$ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}$$

Also $${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$$

$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$

$$ \Rightarrow F{B^2} + F'{B^2} = FF{'^2}$$

$$\therefore$$ $${\left( {\sqrt {{a^2}{e^2} + {b^2}} } \right)^2} + \left( {\sqrt {{a^2}{e^2} + {b^2}} } \right) = {\left( {2ae} \right)^2}$$

$$ \Rightarrow 2\left( {{a^2}{e^2} + {b^2}} \right) = 4{a^2}{e^2}$$

$$ \Rightarrow {e^2} = {{{b^2}} \over {{a^2}}}$$

Also $${e^2} = 1 - {b^2}/{a^2} = 1 - {e^2}$$

$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$

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Complex Numbers

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