Question

Prove that the chord of the parabola $y^2=4ax,$ whose equation is $y-x\sqrt{2}+4a\sqrt{2}=0,$ is a normal to the curve and that its length is $6\sqrt{3}a.$

Answer 1

A line $y=mx+c$ is normal to $y^2=4ax$ if

$c=-2am-a{m}^3$

Given line is $y-x\sqrt{2}+4a\sqrt{2}=0$

$y=x\sqrt{2}-4a\sqrt{2}$

Slope of line $=m=\sqrt{2}$

$c=-2a\sqrt{2}-a{\left(\sqrt{2}\right)}^3=-2a\sqrt{2}-2a\sqrt{2}=-4a\sqrt{2}$

which is same as in the equation of line, hence proved that the chord is normal to the parabola

Now, $y=x\sqrt{2}-4a\sqrt{2}$

Substituting $y$ in the equation of parabola, we get

${(x\sqrt{2}-4a\sqrt{2})}^2=4ax2x^2+32a^2-16ax=4ax2x^2+32a^2-20ax=0x^2-10ax+16a^2=0x^2-8ax-2ax+16a^2=0x(x-8a)-2a(x-8a)=0(x-2a)(x-8a)=0\Rightarrow x=2a,8ay=x\sqrt{2}-4a\sqrt{2}\Rightarrow y=-2a\sqrt{2},4a\sqrt{2}$

So, the point of intersection of chord with the parabola are $A(2a,-2a\sqrt{2})$ and $B(8a,4a\sqrt{2})$

$AB=\sqrt{{(8a-2a)}^2+{(4a\sqrt{2}+2a\sqrt{2})}^2}AB=\sqrt{36a^2+72a^2}=6a\sqrt{3}=6\sqrt{3}a$