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

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Solution

Let P(h,k) be the point of intersection of two
perpendicular tangents equation of pair of tangents is $S S_{1} = T^{2} \Rightarrow (\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1) (\frac{h^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} - 1)$
$= {(\frac{h x}{a^{2}} + \frac{k y}{b^{2}} - 1)}^{2} \Rightarrow \frac{x^{2}}{a^{2}} (\frac{k^{2}}{b^{2}} - 1) + \frac{y^{2}}{a^{2}} (\frac{h^{2}}{a^{2}} - 1) + . . . . . . = 0$ ....(i)
Shince equation (i) represents two perpendicular lines
$∴ \frac{1}{a^{2}} (\frac{k^{2}}{b^{2}} - 1) + \frac{1}{b^{2}} (\frac{h^{2}}{a^{2}} - 1) = 0 \Rightarrow k^{2} - b^{2} + h^{2} - a^{2} = 0 \Rightarrow$ locus is

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