Question

If the distance between a tangent to the parabola $y^2=4x$ and a parallel normal to the same parabola is $2\sqrt{2}$, then possible values of gradient of either of them are:

• A. $-1$
• B. $+1$
• C. $-\sqrt{\sqrt{5}-2}$
• D. $+\sqrt{\sqrt{5}-2}$

Answer 1

Answer: (A), (B)

The correct options are
A $-1$
B $+1$

$y=\frac{x}{t}+aty=-xs+2as+a{s}^{3}s=-\frac{1}{t}$

For parallel condition.

$ts=-1$

$\left|\frac{(2as+a{s}^3)-\left(at\right)}{\sqrt{1+s^2}}\right|=2\sqrt{2}\left|\frac{(2as+a{s}^3)+\frac{a}{s}}{\sqrt{1+s^2}}\right|=\left|\frac{a(s^4+2s^2+1)}{s\sqrt{1+s^2}}\right|=\left|\frac{a{(s^2+1)}^2}{s\sqrt{1+s^2}}\right|=\left|\frac{a{(s^2+1)}^{\frac{3}{2}}}{s}\right|$

As $a=1$

$\left|\frac{{(s^2+1)}^{\frac{3}{2}}}{s}\right|=2\sqrt{2}=2^{\frac{3}{2}}$

Putting $s$ as $\pm 1$ the above equation satisfies.

Hence, the answer is $1,-1$.