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Q.
The locus of point of intersection of two tangents of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, meeting at right angles is
Conic Sections
Solution:
The equation of two perpendicular tangents are
$y = mx +\sqrt{ a ^{2} m ^{2} b ^{2}} \,\,\,\,\,\,...(1)$
and $y =-\frac{1}{ m } x +\sqrt{ a ^{2}\left(-\frac{1}{ m }\right)^{2}+ b ^{2}}$
or $my =- x +\sqrt{ a ^{2}+ b ^{2} m ^{2}}$
Since both tangents will pass through $T ( h , k )$ so
$k - mh =\sqrt{ a ^{2} m ^{2}+ b ^{2}}$ and $mk + h =\sqrt{ a ^{2}+ b ^{2} m ^{2}}$
Squaring and adding
$( k - mh )^{2}+( mk + h )^{2}= a ^{2} m ^{2}+ b ^{2}+ a ^{2}+ b ^{2} m ^{2}$
or $h ^{2}\left( m ^{2}+1\right)+ k ^{2}\left(1+ m ^{2}\right)= a ^{2}\left(1+ m ^{2}\right)+ b ^{2}\left(1+ m ^{2}\right)$
Since $1+m^{2} \neq 0 ; h^{2}+k^{2}=a^{2}+b^{2}$.
Thus, the locus of $T ( h , k )$ is $x ^{2}+ y ^{2}= a ^{2}+ b ^{2}$
This is the equation of a circle. This circle is called the DIRECTOR CIRCLE of the elipse.
