Question

Show that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.

Answer 1

The equation of tangent in terms of slope of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is-

$y=mx+\sqrt{a^2{m}^2+b^2}$

$\Rightarrow y-mx=\sqrt{a^2{m}^2+b^2}$

Squaring both sides, we get

$y^2+m^2{x}^2-2mxy=a^2{m}^2+b^2.....\left(1\right)$

Equation of line perpendicular to the tangent and passes through $(\pm ae,0)$ is-

$my+x=\pm ae$

Squaring both side, we get

$m^2{y}^2+x^2+2mxy=a^2{e}^2.....\left(2\right)$

Adding equation $\left(1\right)\&\left(2\right)$, we have

$(y^2+m^2{x}^2-2mxy)+(m^2{y}^2+x^2+2mxy)=(a^2{m}^2+b^2)+a^2{e}^2$

$(1+m^2)y^2+(1+m^2)x^2=a^2{m}^2+a^2$

$(y^2+x^2)(1+m^2)=a^2{m}^2+a^2$

$(y^2+x^2)(1+m^2)=a^2(1+m^2)$

$\Rightarrow x^2+y^2=a^2\Rightarrow$ Circle

Hence proved that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.