JEE MAIN - Mathematics (2024 - 27th January Morning Shift - No. 4)
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 :
16
24
20
36
Explanation
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 $$\mathrm{C}=(2,8)$$
$$\begin{aligned} & \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \\ & \mathrm{~d}^2=20 \end{aligned}$$
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