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 :
Solution
<p>Equation of normal to parabola</p>
<p>$\mathrm{y=m x-2 m-m^3}$</p>
<p>this normal passing through center of circle $(2,8)$</p>
<p>$$\begin{aligned}
& 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3 \\
& \mathrm{~m}=-2
\end{aligned}$$</p>
<p>So point $\mathrm{P}$ on parabola $\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)$</p>
<p>And $\mathrm{C}=(2,8)$</p>
<p>$$\begin{aligned}
& \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \\
& \mathrm{~d}^2=20
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Conic Sections · Topic: Parabola
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