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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the locus of feet of perpendicular from the foci on and tangent to an ellipse?

$\begin{array}{1 1}(a)\;x^2+y^2=a^2\\(b)\;x+y=a^2\\(c)\;2x^2+3y^2=a^2\\(d)\;\text{None of these}\end{array}$

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The equation of tangent in terms of slope (m) of the ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is
$y=mx+\sqrt{a^2m^2+b^2}$
$y-mx=\sqrt{a^2m^2+b^2}$------(1)
Equation of perpendicular line (1) and passes through ($\pm ae,0)$ is
$my+x=\pm ae$-----(2)
The locus of point of intersection of the line given by equation (1) & (2) can be obtained by eliminating m between them.
Squaring & adding (1) & (2) we get
$y^2(1+m^2)+x^2(1+m^2)=a^2m^2+b^2+a^2e^2$
$(1+m^2)(x^2+y^2)=a^2(m^2+1)$
$x^2+y^2=a^2$
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1
 

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