Question

The normal to the parabola $ y^2 = 8x $ at the point $ (2,4) $ meets the parabola again at the point

• A. (-18, -12)
• B. (-18, 12)
• C. (18, 12)
• D. (18, -12)

Answer 1

Answer: (D)

Given equation is $y^{2}=8 x ....$(i)
$ 2 y \frac{d y}{d x} =8 \Rightarrow \frac{d y}{d x}=\frac{4}{y} $
$\Rightarrow \left(\frac{d y}{d x}\right)_{(2,4)} =1$
$\therefore$ The equation of normal at the point $(2,4)$
and slope $-1$, is $y-4=-1(x-2)$
$\Rightarrow x+y-6=0 ....$ (ii)
$\therefore$ This line intersect the Eq. (i). Solving Eqs.
(i) and (ii), we get
$(6-x)^{2} =8 x$
$\Rightarrow x^{2}+36-12 x =8 x $
$\Rightarrow x^{2}-20 x+36 =0 $
$\Rightarrow x =2,18$
Point $x=2$ is already taken, now
$x=18 \Rightarrow y=\pm 12$
Here, we take $y=-12$ because second point will be below the $X$-axis.
$\therefore$ Required point is $(18,-12)$.