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1

### AIEEE 2012

STATEMENT-1 : An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\sqrt 3$$

STATEMENT-2 :If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x}$$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$

A
Statement-1 is false, Statement-2 is true.
B
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D
Statement-1 is true, Statement-2 is false.

## Explanation

Given equation of ellipse is $$2{x^2} + {y^2} = 4$$

$$\Rightarrow {{2{x^2}} \over 4} + {{{y^2}} \over 4} = 1 \Rightarrow {{{x_2}} \over 2} + {{{y^2}} \over 4} = 1$$

Equation of tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$ is

$$y = mx \pm \sqrt {2{m^2} + 4} \,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

( as equation of tangent to the ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

is $$y=mx+c$$ where $$c = \pm \sqrt {{a^2}{m^2} + {b^2}}$$ )

Now, Equation of tangent to the parabola

$${y^2} = 16\sqrt 3 x$$ is $$y = mx + {{4\sqrt 3 } \over m}\,\,\,\,\,\,\,\,...\left( 2 \right)$$

( as equation of tangent to the parabola

$${y^2} = 4ax$$ is $$y = mx + {a \over m}$$ )

On comparing $$(1)$$ and $$(2),$$ we get

$${{4\sqrt 3 } \over m} = \pm \sqrt {2{m^2} + 4}$$

Squaring on both the sides, we get

$$16\left( 3 \right) = \left( {2{m^2} + 4} \right){m^2}$$

$$\Rightarrow 48 = {m^2}\left( {2{m^2} + 4} \right) \Rightarrow 2{m^4} + 4{m^2} - 48 = 0$$

$$\Rightarrow {m^4} + 2{m^2} - 24 = 0 \Rightarrow \left( {{m^2} + 6} \right)\left( {{m^2} - 4} \right) = 0$$

$$\Rightarrow {m^2} = 4$$ ( as $${m^2} \ne - 6$$ ) $$\Rightarrow m = \pm 2$$

$$\Rightarrow$$ Equation of common tangents are $$y = \pm 2x \pm 2\sqrt 3$$

Thus, statement - $$1$$ is true.

Statement - $$2$$ is obviously true.
2

### AIEEE 2011

Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}}$$ is
A
$$5{x^2} + 3{y^2} - 48 = 0$$
B
$$3{x^2} + 5{y^2} - 15 = 0$$
C
$$5{x^2} + 3{y^2} - 32 = 0$$
D
$$3{x^2} + 5{y^2} - 32 = 0$$

## Explanation

Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$\Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
3

### AIEEE 2010

If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is
A
$$2x+1=0$$
B
$$x=-1$$
C
$$2x-1=0$$
D
$$x=1$$

## Explanation

The locus of perpendicular tangents is directrix

i.e., $$x=-1$$
4

### AIEEE 2009

The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :
A
$${x^2} + 12{y^2} = 16$$
B
$$4{x^2} + 48{y^2} = 48$$
C
$$4{x^2} + 64{y^2} = 48$$
D
$${x^2} + 16{y^2} = 16$$

## Explanation

The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$

So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$

If $$PQRS$$ is the rectangular in which it is inscribed, then

$$P = \left( {2,1} \right).$$

Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

be the ellipse circumscribing the rectangular $$PQRS$$.

Then it passes through $$P\,\,(2,1)$$

$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$

Also, given that, it passes through $$(4,0)$$

$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$

$$\Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$

$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$

or $${x^2} + 12y{}^2 = 16$$

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