Question

Locus of feet of perpendiculars from foci on any tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

• A. $x^2+y^2=9$
• B. $x^2+y^2=16$
• C. $x^2+y^2=25$
• D. ${(x^2+y^2)}^2=16x^2+9y^2$

Answer 1

The correct option is B $x^2+y^2=16$

The focus of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $(ae,0)$ where $e=\sqrt{\frac{a^2-b^2}{a^2}}$

The focus of the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is $(\sqrt{7},0)$

The tangent to an ellipse at $(a\cos \theta ,b\sin \theta )$ can be written as $bx\cos \theta +ay\sin \theta =ab$

The equation here would be $3x\cos \theta +4y\sin \theta =12$ ...(1)

A line passing through the focus, perpendicular to the tangent is $3y\cos \theta -4x\sin \theta =-4\sqrt{7}\sin \theta$ ...(2)

Squaring and adding the equations 1 & 2, we get

$(x^2+y^2)\times (9{\cos }^2\theta +16{\sin }^2\theta )=144+112{\sin }^2\theta$

$\Rightarrow (x^2+y^2)\times (9+7{\sin }^2\theta )=144+112{\sin }^2\theta$

$\Rightarrow x^2+y^2=16$ is the required locus.