First slide

Parabola

Question

Point on $y^{2} = 4 x$ that is nearest to the circle $x^{2} + {(y - 12)}^{2} = 1,$ is

Easy

A

(4,-4)
B

(4, 4)
C

(9, 6)
D

(9, -6)

Solution

Let $P \equiv (t^{2}, 2 t)$
Equation of normal at P is

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

It should pass through (0, 12)

Thus $t^{3} + 2 t - 12 = 0$ $\Rightarrow t = 2$

Hence, $P \equiv (4, 4)$

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