Medium MCQ +4 / -1 PYQ · JEE Mains 2024

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 :

  1. A 16
  2. B 24
  3. C 20 Correct answer
  4. D 36

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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