$P Q$ is a double ordinate of the parabola $y^{2} = 4 a x$ . If the normal at $P$ intersect the line passing through $Q$ and parallel to axis of $x$ at $G$ , then locus of $G$ is a parabola with

length of latus rectum equal to

4 a

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vertex at

(4 a, 0)

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directrix as the line

x - 3 a = 0

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focus

(5 a, 0)

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Solution

The correct options are
A length of latus rectum equal to $4 a$
B vertex at $(4 a, 0)$
C directrix as the line $x - 3 a = 0$
D focus $(5 a, 0)$
$y^{2} = 4 a x$

$2 y y^{'} = 4 a$

$y^{'} = \frac{2 a}{y} \Rightarrow y^{'} = \frac{2 a}{k}$

(Normal) $m_{n} = \frac{- k}{2 a}$ equation of line 2

$y - k = \frac{- k}{2 a} (x - h)$

$2 a y - 2 a k = - k x + k h$

$- 4 a k + k 4 - k h = 0$

$k [- 4 a + x - h] = 0$

$x = h + 4 a \Rightarrow x - 4 a$

$y - - k$

$k^{2} = 4 a h$

$y^{2} = 4 a (x - 4 a)$

Vertex $= (4 a, 0)$ and focus $(5 a, 0)$

$L R = 4 a$

$x = 4 a - a$

$x - 3 a = 0$

All option are cottrct

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