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The locus of the point of intersection of perpendicular tangents to the hyperbola $\large\frac{x^{2}}{16}-\frac{y^{2}}{9}$=$1$ is

\[\begin{array}{1 1} (1)x^{2}+y^{2}=25&(2)x^{2}+y^{2}=4\\(3)x^{2}+y^{2}=3&(4)x^{2}+y^{2}=7\end{array}\]

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The equation of the hyperbola is
$\large\frac{x^2}{16} -\frac{y^2}{9}$$=1$
$a^2=16,b^2=9$
Required locus is $ x^2+y^2=16-9$
$x^2+y^2=7$
Hence 4 is the correct answer.
answered May 16, 2014 by meena.p
 

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