Question

If the normals of the parabola $y^{2}=4x$ drawn at the end points of its latus rectum are tangents to the circle $\left(x - 3\right)^{2}+\left(y + 2\right)^{2}=r^{2}$ , then the value of $r^{4}$ is equal to

Answer 1

End points of the latus rectum of the parabola $y^{2}=4x$ and $\left(1 , \pm 2\right).$
Equation of the normals at points $\left(1 , \pm 2\right)$ are
$y=-x+3$ and $y=x-3$
rr $x+y-3=0$ and $x-y-3=0$
These lines are tangent to the circle $\left(x - 3\right)^{2}+\left(y + 2\right)^{2}=r^{2}$
$\therefore \left|\frac{3 \pm 2 - 3}{\sqrt{1 + 1}}\right|=r=\sqrt{2}$
Hence, $r^{4}=4$