Question

Find the lengths of the normals drawn from the point on the axis of the parabola $y^2=8ax$ whose distance from the focus is $8a.$

Answer 1

For $y^2=8ax$

Focus is $(2a,0)$

Point on axis whose distance from focus is $8a$ is $P(10a,0)$

Equation of normal is

$y=mx-2\left(2a\right)-\left(2a\right)m^3$

$y=mx-4a-2a{m}^3$

It passes through $P$

$\therefore 0=m\left(10a\right)-4am-2a{m}^3$

$2a{m}^3-6am=02m(m^2-3)=0$

$\Rightarrow m=0,\pm \sqrt{3}$

For $m=0$ equation of normal is

$y=0$

So the length of normal is $10a$

For $m=\sqrt{3}$

Equation of normal is

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

$y=\sqrt{3}x-10\sqrt{3}a$

at $y=0$

$0=\sqrt{3}x-10\sqrt{3}a$

$x=10a$

$\Rightarrow N(10a,0)$

Feet of normal is

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

$\Rightarrow P(6a,-4a\sqrt{3})$

Length of normal $=PN$

$PN=\sqrt{{(6a-10a)}^2+{(-4a\sqrt{3}-0)}^2}$

$PN=8a$

For $m=-\sqrt{3}$ the normal is symmetric so its length will also be $8a$

So the lengths of normal are $10a,8a,8a$