Question

The locus of the foot of perpendiculars drawn from the vertex on a variable tangent to the parabola $y^2=4x$ is

• A. $x(x^2+y^2)+x^2=0$
• B. $y(x^2+y^2)+x^2=0$
• C. $y(x^2+y^2)+y^2=0$
• D. $x(x^2+y^2)+y^2=0$

Answer 1

The correct option is D $x(x^2+y^2)+y^2=0$

Let any tangent to the parabola $y^2=4x$ be $ty=x+t^2$ and the foot of perpendicular from the vertex $(0,0)$ to this tangent be $(h,k)$
$\frac{h-0}{1}=\frac{k-0}{-t}=\frac{-t^2}{1+t^2}$
$\Rightarrow h=\frac{-t^2}{1+t^2}...\left(1\right)$
$\Rightarrow k=\frac{t^3}{1+t^2}...\left(2\right)$
dividing equation $\left(1\right)$ with equation $\left(2\right)$
$\frac{h}{k}=\frac{-1}{t}\left(3\right)$
from equation $\left(1\right)$ and $\left(3\right)$
$h=\frac{-(-\frac{k}{h}{)}^2}{1+(-\frac{k}{h}{)}^2}=\frac{-k^2}{h^2+k^2}$
$\Rightarrow h(h^2+k^2)+k^2=0$
replacing $(h,k)$ with $(x,y)$
we get
$x(x^2+y^2)+y^2=0$