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 $y^2=8x....$(i)
$2y\frac{dy}{dx}=8\Rightarrow \frac{dy}{dx}=\frac{4}{y}$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,4)}=1$
$\therefore$ The equation of normal at the point $(2,4)$
and slope $-1$, is $y-4=-1(x-2)$
$\Rightarrow x+y-6=\mathrm{0....}$ (ii)
$\therefore$ This line intersect the Eq. (i). Solving Eqs.
(i) and (ii), we get
$(6-x{)}^2=8x$
$\Rightarrow x^2+36-12x=8x$
$\Rightarrow x^2-20x+36=0$
$\Rightarrow x=2,18$
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 $(18,-12)$.