Parabola Question 1
Question 1 - 2024 (27 Jan Shift 1)
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) 16
(2) 24
(3) 20
(4) 36
Show Answer
Answer (3)
Solution
Equation of normal to parabola
$y=m x-2 m-m^{3}$
this normal passing through center of circle $(2,8)$
$8=2 m-2 m-m^{3}$
$m=-2$
So point $\mathrm{P}$ on parabola $\Rightarrow\left(\mathrm{am}^{2},-2 \mathrm{am}\right)=(4,4)$
And $\mathrm{C}=(2,8)$
$\mathrm{PC}=\sqrt{4+16}=\sqrt{20}$
$\mathrm{d}^{2}=20$
