Question

Find the locus of point of intersection of perpendicular tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Answer 1

Let P (h, k) be the point of intersection of two perpendicular tangents. Equation of pair of tangents is $S{S}_1=T^2\Rightarrow (\frac{x^2}{a^2}-\frac{y^2}{b^2}-1)(\frac{h^2}{a^2}-\frac{k^2}{b^2}-1)={(\frac{hx}{a^2}-\frac{ky}{a^2}-1)}^2\Rightarrow \frac{x^2}{a^2}(-\frac{k^2}{b^2}-1)-\frac{y^2}{b^2}(\frac{h^2}{a^2}-1)+......=0$

Since equation (i) represents two perpendicular lines

$\therefore \frac{1}{a^2}(-\frac{k^2}{b^2}-1)-\frac{1}{b^2}(\frac{h^2}{a^2}-1)=0\Rightarrow -k^2-b^2-h^2+a^2=0\Rightarrow$ locus is $x^2+y^2=a^2-b^2$