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1

### AIEEE 2004

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

### AIEEE 2003

The foci of yhe ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is
A
$$9$$
B
$$1$$
C
$$5$$
D
$$7$$

## Explanation

$${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$

$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$

$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$

$$\therefore$$ Foci $$= \left( { \pm 3,0} \right)$$

$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola

$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$

$$\therefore$$ $$e = {3 \over 4}$$

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

$$\Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$
3

### AIEEE 2003

The normal at the point$$\left( {bt_1^2,2b{t_1}} \right)$$ on a parabola meets the parabola again in the point $$\left( {bt_2^2,2b{t_2}} \right)$$, then
A
$${t_2} = {t_1} + {2 \over {{t_1}}}$$
B
$${t_2} = -{t_1} - {2 \over {{t_1}}}$$
C
$${t_2} = -{t_1} + {2 \over {{t_1}}}$$
D
$${t_2} = {t_1} - {2 \over {{t_1}}}$$

## Explanation

Equation of the normal to a parabola $${y^2} = 4bx$$ at point

$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$

As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then

$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$

$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$

$$= - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$

$$\Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$

$$\Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$

$$\Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$
4

### AIEEE 2002

Two common tangents to the circle $${x^2} + {y^2} = 2{a^2}$$ and parabola $${y^2} = 8ax$$ are
A
$$x = \pm \left( {y + 2a} \right)$$
B
$$y = \pm \left( {x + 2a} \right)$$
C
$$x = \pm \left( {y + a} \right)$$
D
$$y = \pm \left( {x + a} \right)$$

## Explanation

Any tangent to the parabola $${y^2} = 8ax$$ is

$$y = mx + {{2a} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

If $$(i)$$ is a tangent to the circle, $${x^2} + {y^2} = 2{a^2}$$ then,

$$\sqrt {2a} = \pm {{2a} \over {m\sqrt {{m^2} + 1} }}$$

$$\Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2$$

$$\Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$

$$\Rightarrow m = \pm 1.$$

So from $$(i),$$ $$y = \pm \left( {x + 2a} \right).$$

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