If the shortest distance of the parabola from the centre of the circle is d, then is equal to : [2024]

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 d, then $d^{2}$ is equal to : [2024]

Smart Achievers · Accepted Answer

(3)

Circle $x^{2} + y^{2} - 4 x - 16 y + 64 = 0$

Centre $\equiv$ (2, 8), Radius = 2 units

Parabola : $y^{2} = 4 x$

a = 1

Equation of normal at $(t^{2}, 2 t)$ of parabola:

$x t + y = a t^{3} + 2 a t$

$x t + y = t^{3} + 2 t$ ... (i)

Since, normal passes through centre, then

$2 t + 8 = t^{3} + 2 t$ ... (ii)

$\Rightarrow t^{3} = 8$

$\Rightarrow t^{3} - 8 = 0$

$\Rightarrow (t - 2) (t^{2} + 4 + 2 t) = 0$

$\Rightarrow t = 2 (∵ t \geq 0)$

d = distance between (2, 8) and (4, 4) = $\sqrt{4 + 16} = \sqrt{20}$

$\Rightarrow d^{2} = 20$ .

Solution

Q.
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 d, then $d^{2}$ is equal to : [2024]

1 16

2 24

3 20

4 36

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