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KEAM 2011 Math Paper

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Question : 16 of 120

Marks: +1, -0

For each point

(x, y)

on an ellipse, the sum of the distances from

(x, y)

(2, 0)

(-2, 0)

is 8. Then the positive value of

x

(x, 3)

lies on the ellipse is

$2$
$2\sqrt{3}$
$\frac{1}{\sqrt{3}}$
4
0

Solution:

Given that,

PS+PS'=8

PS=8-PS'

Squaring on both sides,

(PS)^2=64+(PS')^2-16(PS')

\{(x-2)^2+y^2\}=64+\{(x+2)^2+y^2\}

-16\sqrt{(x+2)^2+y^2}

[ \because PS=\sqrt{(x+2)^2+y^2} ]

[ \because PS=\sqrt{(x-2)^2+y^2} ]

(x^2+y^2+4-4x)=(x^2+y^2+4+4x+64)-16\sqrt{x^2+y^2+4x+4}

\Rightarrow \;\; 16\sqrt{x^2+y^2+4+4x}=8x+64

\Rightarrow \;\; 2\sqrt{x^2+y^2+4x+4}=(x+8)

Squaring on both sides;

4x^2+4y^2+16x+16=x^2+64+16x

\Rightarrow 3x^2+4y^2=48,

which is equation of an ellipse given

(y=3), \because (x,3)

lies on an ellipse.

\Rightarrow \;\; 3x^2+4\times 9=48

\Rightarrow \;\; 3x^2=12

\Rightarrow \;\; x^2=4

(here we take only positive value of x. )

\Rightarrow\; x=\pm 2

\Rightarrow\; x=2

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