First slide

Parabola

Question

Normal at a point $P$ on the parabola $y^{2} = 4 a x$ meets
the axis at $Q$ such that the distance of $Q$ from the
focus of the parabola is 10 $a$ . The coordinates of $P$
are:

Moderate

A

$(6 a, 9 a)$
B

$(6 a, - 9 a)$
C

$(9 a, 6 a)$
D

$(3 a, 6 a)$

Solution

Equation of the normal at $P (a t^{2}, 2 a t)$ to the parabola
$y^{2} = 4 a x is y = - t x + 2 a t + a t^{3}$ which meets the
axis $y = 0$ at $x = a (2 + t^{2})$ . So coordinates of $Q$ are
$(a (2 + t^{2}), 0) .$ Focus is $S (a, 0)$
$Q S = 2 a + a t^{2} - a = a (1 + k^{2}) = 10 a \Rightarrow t^{2} = 9$
So the required coordinates are $(9 a, \pm 6 a)$ .

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