Question

What normal to curve $y=x^2$ forms the shortest chord ?

Answer 1

First, note that the derivative of $y=x^{2}is{y}^{\prime }=2x$, which means that the slope of the tangent at $(x,x^2)$ is $2x$ so the slope of the normal at $(x,x^2)is-1/2x.$

An equation of the normal at $(a,a^2)$ is
$y-a^2=-1/\left(2a\right)(x-a)$
Observe that the line intersection $y=x^2$ at the solutions of the system
$y=x^2,y-a^2=-1/\left(2a\right)(x-a):$
$(a,a^2)and(-1/(2a)-a,1+1/(4a^2)+a^2).$
Differentiating gives us

$l^{\prime }\left(a\right)=(-1-6a^2+32a^6)/\left(4a^5\right)$
$=(2a^1-1)(4a^2+1{)}^2/(4a^5)$
and solving
$l^{\prime }\left(a\right)=0givesusa=-1/\sqrt{2}ora=1/\sqrt{2}.$