Question 1076007
{{{3ay^2=x^2(x+a)}}}
Implicitly differentiate both sides,
{{{6aydy=x^2dx+(x+a)(2xdx)}}}
{{{6aydy=(x^2+2x(x+a))dx}}}
{{{dy/dx=(x^2+2x(x+a))/(6ay)}}}
When {{{x=2a}}},
{{{dy/dx=((2a)^2+2(2a)(2a+a))/(6a(2a))}}}
{{{dy/dx=(4a^2+12a^2)/(12a^2)}}}
{{{dy/dx=(16a^2)/(12a^2)}}}
{{{dy/dx=4/3}}}
The normal is perpendicular to the tangent and the slopes are negative reciprocals,
{{{m[N]*(4/3)=-1}}}
{{{m[N]=-3/4}}}
So then using the point-slope form,
{{{y-2a=-(3/4)(x-2a)}}}
{{{4(y-2a)=-3(x-2a)}}}
{{{4y-8a=-3x+6a}}}
{{{highlight(3x+4y=14a)}}}