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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

Find the equation of an ellipse whose focus is $(-1,1)$ eccentricity is $\large\frac{1}{2}$ and the directix is $x-y+3=0?$

$\begin{array}{1 1}(a)\;7x^2+7y^2+2xy+10x-10y+7=0\\(b)\;7x^2-7y^2-2xy+10x-10y-7=0\\(c)\;7x^2+7y^2+2xy+10x+10y+10=0\\(d)\;x^2+y^2+2xy+10x-10y+7=0\end{array}$

1 Answer

SP=ePM
$(SP)^2=e^2(PM)^2$
$(x+1)^2+(y-1)^2=\large\frac{1}{4}\{\large\frac{x-y+3}{\sqrt{2}}\}^2$
$\Rightarrow 8(x^2+y^2+2x-2y+2)=x^2+y^2+9-2xy+6(x-y)$
$\Rightarrow 7x^2+7y^2+2xy+10x-10y+7=0$
Which is the required equation of ellipse.
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
 

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