Question

If the normals at P and Q meet on the parabola, prove that the point of intersection of the tangents at P and Q lies either on a certain straight line, which is parallel to the tangent at the vertex, or on the curve whose equation is $y^2(x+2a)+4a^3=0$.

Answer 1

A normal to the parabola at a point $(a{t}^2,2at)$ is given by $y+tx=2at+a{t}^3$

The intersection of normals at points $t_1,t_2$ is given by $(2a+a(t_1^2+t_2^2+t_1{t}_2),-a{t}_1{t}_2(t_1+t_2\left)\right)$

Since this point lies on the parabola, it satisfies the equation.

$\therefore (-a{t}_1{t}_2(t_1+t_2){)}^2=4a\times (2a+a(t_1^2+t_2^2+t_1{t}_2))$

The point of intersection of tangents is given by $x=a{t}_1{t}_2,y=a(t_1+t_2)$

$\Rightarrow$ we get $x^2\times \frac{y^2}{a^2}=8a^2+4\times (y^2-ax)$

i.e. $x^2{y}^2=8a^4+4a^2{y}^2-4a^{3}x$

i.e. $y^2(x^2-4a^2)=8a^3(a-\frac{x}{2})$

i.e. $y^2(x+2a)(x-2a)=4a^3(2a-x)$

i.e. $(x-2a)\left(y^2\right(x+2a)+4a^3)=0$

$\therefore$ either $x=2a$ or $y^2(x+2a)+4a^3=0$