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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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The equation of the ellipse whose focus is $(1,-1)$,the directrix the line $x-y-3=0$ and eccentricity $\large\frac{1}{2}$ is

$\begin{array}{1 1}7x^2+2xy+7y^2+10x-10y+7=0\\7x^2+2xy+7y^2+7=0\\7x^2+2xy+7y^2+10x-10y-7=0\\7x^2-2xy-7y^2-7=0\end{array} $

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  • For an ellipse the ratio $\large\frac{SP}{PM}$$=e$ where $e < 1$ and $S(ae,0)$ is the focus and $P(x,y)$ is any point on the ellipse and $PM$ is the perpendicular to the directrix.
Answer : $7x^2+2xy+7y^2+10x-10y+7=0$
Given focus is $(1,-1)$ and equation of the directrix is $x-y=3$
Eccentricity $e=\large\frac{1}{2}$
Hence $SP=ePM$
$SP^2=e^2PM^2$
(ie) $SP^2=\large\frac{1}{4}$$(PM)^2$
$4SP^2=PM^2$
(ie) $4[(x+1)^2+(y-1)^2]=\big(\large\frac{x-y-3}{\sqrt{1^2+(-1)^2}}\big)^2$
$\Rightarrow 8(x^2+y^2+2x-2y+2)=(x-y+3)^2$
On simplifying we get,
$7x^2+2xy+7y^2+10x-10y+7=0$
 
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The equation of ellipse whose focus is (-1,1) the directrix the line x-y+3=0 and eccentricity 1 by 2
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