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1

### AIEEE 2005

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

## Explanation

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.
2

### AIEEE 2005

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 $$PQ$$

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

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

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

$$\Rightarrow {y^2} - 4x + 2 = 0.$$
3

### AIEEE 2005

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

$$\Rightarrow 2{e^2} = 1,\,\,e = {1 \over {\sqrt 2 }}$$
4

### AIEEE 2004

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

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

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