Question

The locus of the foot of the perpendicular drawn from the center upon any tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ is

• A. $(x^2+y^2{)}^2=16x^2+9y^2$
• B. $(x^2+y^2{)}^2=9x^2+16y^2$
• C. $x^2+y^2=25$
• D. $x^2+y^2=144$

Answer 1

The correct option is A $(x^2+y^2{)}^2=16x^2+9y^2$

Let the foot of the pependicular be $P(h,k)$
Center is $O(0,0)$
So, slope of $OP$ is $\frac{k}{h}$
$\therefore$ Slope of the tangent is $\frac{-h}{k}$
So the equation of the tangent at $P(h,k)$ is
$y-k=-\frac{h}{k}(x-h)\Rightarrow yk+xh=k^2+h^2\Rightarrow y=(-\frac{h}{k})x+\frac{h^2+k^2}{k}$
Since it touches the ellipse,
$\therefore \frac{h^2+k^2}{k}=\sqrt{16\frac{h^2}{k^2}+9}\Rightarrow (x^2+y^2{)}^2=16x^2+9y^2$