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1

AIEEE 2005

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

Explanation

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

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

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

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

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

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

AIEEE 2005

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

Explanation

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

$$ \Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$
3

AIEEE 2004

MCQ (Single Correct Answer)
The eccentricity of an ellipse, with its centre at the origin, is $${1 \over 2}$$. If one of the directrices is $$x=4$$, then the equation of the ellipse is:
A
$$4{x^2} + 3{y^2} = 1$$
B
$$3{x^2} + 4{y^2} = 12$$
C
$$4{x^2} + 3{y^2} = 12$$
D
$$3{x^2} + 4{y^2} = 1$$

Explanation

$$e = {1 \over 2}.\,\,$$ Directrix, $$x = {a \over e} = 4$$

$$\therefore$$ $$a = 4 \times {1 \over 2} = 2$$

$$\therefore$$ $$b = 2\sqrt {1 - {1 \over 4}} = \sqrt 3 $$

Equation of elhipe is

$${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1 \Rightarrow 3{x^2} + 4{y^2} = 12$$
4

AIEEE 2004

MCQ (Single Correct Answer)
If $$a \ne 0$$ and the line $$2bx+3cy+4d=0$$ passes through the points of intersection of the parabolas $${y^2} = 4ax$$ and $${x^2} = 4ay$$, then
A
$${d^2} + {\left( {3b - 2c} \right)^2} = 0$$
B
$${d^2} + {\left( {3b + 2c} \right)^2} = 0$$
C
$${d^2} + {\left( {2b - 3c} \right)^2} = 0$$
D
$${d^2} + {\left( {2b + 3c} \right)^2} = 0$$

Explanation

Solving equations of parabolas

$${y^2} = 4ax$$ and $${x^2} = 4ay$$

we get $$(0,0)$$ and $$(4a, 4a)$$

Substituting in the given equation of line

$$2bx+3cy+4d=0,$$

we get $$d=0$$

and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$

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