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

Answer 1

Answer: (C)

$C =-29 \,m -9\, m ^{3}$
$a =2$
Given $\left( at ^{2}- a \right)^{2}+4 a ^{2} t ^{2}=64$
$\rightarrow\left( a \left( t ^{2}+1\right)\right)=8$
$\Rightarrow t ^{2}+1=4= t ^{2}=3$
$\Rightarrow t=\sqrt{3}$
$\therefore C=-a\left[-2 t-t^{3}\right]=2 a t\left(2+t^{2}\right)$
$=2 \sqrt{3}(5)$
$|C|=10 \sqrt{3}$