Question

The slope of the shortest normal chord of the curve $y=x^2$ is

• A. $\pm \sqrt{2}$
• B. $\pm \frac{1}{\sqrt{2}}$
• C. $\pm \frac{1}{2\sqrt{2}}$
• D. $\pm \frac{1}{\sqrt{3}}$

Answer 1

Answer: (B)

$\pm \frac{1}{\sqrt{2}}$

solve:

Given, $y=x^2$ is a parabola

it Parametric coordinate

$p(a,a^2)$

slope of tangent, $\frac{dy}{dx}=2x$

$\Rightarrow$ slope of normal $={(-\frac{dx}{dy})}_{x=a}=-\frac{1}{2a}$

eqn. of normal $\Rightarrow (y-a^2)=-\frac{1}{2a}(x-a)$

Lel $Q$ be the point where it intersect

parabola

$\Rightarrow Q(-(\frac{1}{2a}+a),1+\frac{1}{\left(4a^2\right)}+a^2)$

so,length of $PQ,l\left(a\right)=\sqrt{{(-\frac{1}{2a}-a-a)}^2+{(1+\frac{1}{4a^2}+a^2-a^2)}^2}$

$\Rightarrow l^2\left(a\right)=\frac{{(1+4a^2)}^3}{16a^4}$

for, length of normal chord be shortest, $l^2\left(a\right)$ is also shortest and $\frac{d{l}^2\left(a\right)}{da}=0$ at that Point,

$\Rightarrow \frac{d{l}^2\left(a\right)}{da}=\frac{(2a^2-1){(4a^2+1)}^2}{4a^5}=0$

$\Rightarrow a=+\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$

So, slope of shortest normal chord $=\frac{-1}{2a}$

$=\pm \frac{1}{\sqrt{2}}$