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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of director circle for ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$?

$\begin{array}{1 1}(a)\;x^2-y^2=a^2+b^2\\(b)\;x^2+y^2=a^2+b^2\\(c)\;x^2-y^2=a^2-b^2\\(d)\;\text{None of these}\end{array}$

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  • Director circle is locus of point of intersection of $\perp$ tangents to the ellipse.
Let any tangent in terms of slope of ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is
$y=mx+\sqrt{a^2m^2+b^2}$
It passes through (h,k)
$k=mh+\sqrt{(a^2m^2+b^2}$
$(k-mh)^2=a^2m^2+b^2$
$k^2+m^2h^2-2mhk=a^2m^2+b^2$
$m^2(h^2-a^2)-2hkm+k^2-b^2=0$
It is quadratic equation in m let slope of two tangent are $m_1$ and $m_2$
$m_1m_2=\large\frac{k^2-b^2}{h^2-a^2}$
$-1=\large\frac{k^2-b^2}{h^2-a^2}$(tangents are $\perp)$
$h^2+k^2=a^2+b^2$
$x^2+y^2=a^2+b^2$
This is the required equation.
Hence (b) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Mar 24, 2014 by rvidyagovindarajan_1
 

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