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Q. The locus of the point of intersection of perpendicular tangents to the ellipse is called

KCETKCET 2012Conic Sections

Solution:

The equation of a pair of tangents from $\left(\alpha, \beta\right)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$ is
$\left(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-1\right)\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) = \left(\frac{x\alpha}{a^{2}}+\frac{y\beta}{b^{2}}-1\right)^{2}$
$\left(SS_{1} = T^{2}\right)$
The tangents are perpendicular, if the coefficient of $x^{2} +$ the coefficient of $y^{2} =0.$
$\Rightarrow \frac{1}{a^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right)-\frac{\alpha^{2}}{a^{4}}+\frac{1}{b^{2}}\left(\frac{\alpha^{2}}{a^{2}}+\frac{\beta^{2}}{b^{2}}-1\right) -\frac{\beta^{2}}{b^{4}} = 0$
$\Rightarrow \frac{\beta^{2}}{a^{2}b^{2}}-\frac{1}{a^{2}}+\frac{\alpha^{2}}{a^{2}b^{2}}-\frac{1}{b^{2}} = 0$
$\Rightarrow \alpha^{2} + \beta^{2} = a^{2}b^{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)$
$= a^{2}b^{2}\left(\frac{a^{2}+b^{2}}{a^{2}b^{2}}\right)
= a^{2} + b^{2}$
Hence, the locus of $\left(\alpha, \beta\right)$ is the circle
$x^{2} + y^{2} = a^{2} + b^{2}$
This circle is called the director circle.