Match the following loci for the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1(\mathrm{a}>\mathrm{b})$
 

	COLUM N-I		COLUM N-II
A)	Locus of point of intersection of two perpendicular Tangents	P)	${\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2={\mathrm{a}}^2{\mathrm{x}}^2+{\mathrm{b}}^2{\mathrm{y}}^2$
B)	Locus of foot of perpendicular from any focus upon Any tangent	Q)	$4{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^2={\mathrm{a}}^2{\mathrm{x}}^2+{\mathrm{b}}^2{\mathrm{y}}^2$
C)	Locus of foot of the perpendicular from centre on any tangent	R)	${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2$
D)	Locus of mid point of segment OM where M is the foot of the  perpendicular from centre O to any tangent	S)	${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{a}}^2+{\mathrm{b}}^2$

 

The locus of the point of intersection of perpendicular tangents to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is its director circle $x^2+y^2=a^2+b^2.$

Ttangent to the ellipse $y=mx+\sqrt{a^2{m}^2+b^2}$
the above tangent is of the form  $y =-\frac{1}{m} x + c$
passes through the focus S(ae, 0)
$\begin{array}{l}0=-\frac{ae}{m}+c\\ \Rightarrow \mathrm{c}=\frac{\mathrm{ae}}{\mathrm{m}}\end{array}$
$y=\frac{-x}{m}+\frac{ae}{m}\Rightarrow x+my=ae$
$\begin{array}{l}(y-mx{)}^2+(x+my{)}^2\\ =a^2{m}^2+b^2+a^2{e}^2\\ \Rightarrow \left(1+m^2\right)\left(x^2+y^2\right)\left[\because b^2=a^2\left(1-e^2\right)\right]\\ =a^2{m}^2+a^2\\ \Rightarrow x^2+y^2=a^2\end{array}$
$\text{ The}\text{ equation of tangent to ellipse }y=mx+\sqrt{\left(a^2{m}^2+b^2\right)}$..........(1)
is a tangent to the ellipse. The line passing through the centre (0, 0) and perpendicular to the line which is given in Eq.
 x + my = 0   .....(2) From Eqs. (1) and (2), we have
$\begin{array}{r}y=x\left(\frac{-x}{y}\right)+\sqrt{\frac{a^2{x}^2}{y^2}+b^2}\\ \Rightarrow x^2+y^2=\sqrt{a^2{x}^2+b^2{y}^2}\\ \Rightarrow {\left(x^2+y^2\right)}^2=a^2{x}^2+b^2{y}^2\end{array}$