Normals at $P$, $Q$, $R$ on parabola $y^2=4ax$ meet at a point on line $x=k$. Prove that sides of $\triangle PQR$ touch parabola $y^2=16a(x+2a-k)$.
I have tried using the equation of the normal in parametric form, but I did not get any result. I have also used the equation of tangents, and also through transformation of axis of the parabola, but the result was confusing.
We can use basic coordinate geometry to find the equation of the normals. Let $y_p$, $y_q$ and $y_r$ be the y coordinates of the points p, q and r on the parabola. Then $\frac{y_p^2}{4a}$, $\frac{y_q^2}{4a}$ and $\frac{y_r^2}{4a}$ are the x coordinates. Then the equation of the normal at p is $$y = -\frac{y_px}{2a} +\frac{y_p^3}{8a^2} + y_p$$ and the normal at q is $$y = -\frac{y_qx}{2a} +\frac{y_q^3}{8a^2} + y_q$$ Solving these two equations we have $$x=\frac{y_q^2+y_py_q+y_p^2}{4a}+2a$$ This equals k per question. Next we use the coordinates of points p and q to find the equation of $pq$. We have $$x=\frac{(y_p+y_q)y - y_py_q}{4a}$$. Then put this x in the given parabola we have $$y^2 - 4(y_p+y_q)y + 4y_py_q - 16a(2a-k) =0$$
The last step is to prove this quadratic equation has only one root in order to be a tangent. Since we have $$16(y_p+y_q)^2-16y_py_q+64a(2a-k)=0$$ because $$y_q^2+y_py_q+y_p^2=4a(k-2a)$$ the line pq cuts the given parabola at only one point and is a tangent to the second given parabola. This is also true for the the line joining qr and $pr$.
