Question

If $y=mx+c$ is the normal at a point on the parabola $y^2=8x$ whose focal distance is $8$ units, then $|c|$ is equal to.

• A. $2\sqrt{3}$
• B. $10\sqrt{3}$
• C. $8\sqrt{3}$
• D. $16\sqrt{3}$

Answer 1

The correct option is B $10\sqrt{3}$

$y=mx+c$
$y^2=8x$
$=a=2$
$2y{y}^{\prime }=8,\Rightarrow y^{\prime }=\frac{4}{y}$
Focal distance of a point $(2t^2,4t)=2(1+t^2)$
$2(1+1^2)=8$
$1+t^2=4$
$t^2=3$
$t=\pm \sqrt{3}$
Slope of tangent at $\left(2\right(\pm \sqrt{3}{)}^2,4(\pm \sqrt{3})):(6,\pm 4\sqrt{3})=\frac{4}{y}{|}_{y=\pm 4\sqrt{3}}=\pm \frac{1}{\sqrt{3}}$
Slope of normal $=\pm \sqrt{3}$
$m=\pm \sqrt{3}$
$y=\pm \sqrt{3x}+c$
This line passes through $(6,\pm 4\sqrt{3})$
$\therefore \pm 4\sqrt{3}=\pm \sqrt{3}\left(6\right)+c$
$\pm 4\sqrt{3}=\pm 6\sqrt{3}+c$
$\therefore c=\pm 10\sqrt{3}$

$|c|=10\sqrt{3}$