If the shortest distance of the parabola y^2=4 x from the centre of the circle x^2+y^2-4 x…

Question

If the shortest distance of the parabola $y^2=4 x$ from the centre of the circle $x^2+y^2-4 x-16 y+64=0$ is $\mathrm{d}$, then $\mathrm{d}^2$ is equal to :

Accepted Answer

Equation of normal to parabola

$$\mathrm{y=m x-2 m-m^3}$$

this normal passing through center of circle $(2,8)$

$$\begin{aligned}
& 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3 \
& \mathrm{~m}=-2
\end{aligned}$$

So point $$\mathrm{P}$$ on parabola $$\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)$$

And $$\mathrm{C}=(2,8)$$

$$\begin{aligned}
& \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \
& \mathrm{~d}^2=20
\end{aligned}$$

Solutions · 1