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# Find the locus of intersection of two perpendicular tangent to the hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1? \begin{array}{1 1}(a)\;h^2+k^2=a^2-b^2\\(b)\;h^2+k^2=a^2+b^2\\(c)\;h^2-k^2=a^2-b^2\\(d)\;k^2-h^2=a^2-b^2\end{array} Can you answer this question? ## 1 Answer 0 votes Let any tangent in terms of slope of hyperbola \large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is
$y=mx+\sqrt{a^2m^2+b^2}$
It passes through (h,k)
$k=mh+\sqrt{a^2m^2+b^2}$
$(k-mh)^2=a^2m^2-b^2$
$m^2(h^2-a^2)-2hkm+k^2+b^2=0$
Slope of tangent be $m_1$ & $m_2$
$m_1m_2=\large\frac{k^2+b^2}{h^2-a^2}$
$-1=\large\frac{k^2+b^2}{h^2-a^2}$(tangent are $\perp$)
Hence $h^2+k^2=a^2-b^2$
Hence (a) is the correct answer.