Here is my problem: If the normals at the ends of a variable chord PQ of the parabola $y^2-4y-2x=0$ are perpendicular then the tangents at P and Q will intersect at?? The correct answer is $2x+5=0$.
I could not figure out the way from where to start. Basically i am strucked with the equation of parabola since it is not the standard parabola. The vertex of the given parabola is $(-2,2)$. Thanks for any kind of help.
We have $x=(y^2-4y)/2$
If $y=2t,x=2t^2-4t$
So, any point $(A)$ on $2x=y^2-4y$ can be written as $(2t^2-4t,2t)$
The equation of the tangent at $A(2t^2-4t,2t)$ will be $$2t^2-2ty+x+2y=0\ \ \ \ \ (1)$$ which is a Quadratic Equation in $t$
If $P(t_1),Q(t_2)$ $t_1t_2=\dfrac{x+2y}2;t_1+t_2=y\ \ \ \ (2)$
$(1)$ can also be expressed as $y=\dfrac x{2(t-1)}+\dfrac{t^2}{t-1}$
So, the gradient of the normal at $A(2t^2-4t,2t)$ will be $\dfrac{-1}{\dfrac1{2(t-1)}}=2(1-t)$
$\implies2(1-t_1)\cdot2(1-t_2)=-1\iff 1-(t_1+t_2)+t_1t_2=-\dfrac14\ \ \ \ (3)$
Compare $(2),(3)$ to eliminate $t_1,t_2$
Let $P\left(\frac{p^2-4p}{2},p\right),Q\left(\frac{q^2-4q}{2},q\right)$ where $p\not=q$.
Now, one has $x=\frac{y^2-4y}{2}\Rightarrow \frac{dx}{dy}=\frac{2y-4}{2}=y-2$. Since the normals at $P,Q$ are perpendicular, one has $$(p-2)(q-2)=-1\iff pq-2p-2q=-5\tag 1$$
By the way, since the tangents at $P,Q$ are $$y-p=\frac{1}{p-2}\left(x-\frac{p^2-4p}{2}\right)\tag2$$ $$y-q=\frac{1}{q-2}\left(x-\frac{q^2-4q}{2}\right)\tag3$$ respectively, $(2)-(3)$ gives us $$\begin{align}q-p&=\left(\frac{1}{p-2}-\frac{1}{q-2}\right)x-\frac{1}{p-2}\cdot \frac{p^2-4p}{2}+\frac{1}{q-2}\cdot\frac{q^2-4q}{2}\\&=\frac{q-p}{(p-2)(q-2)}x+\frac{(q-p)(pq-2p-2q+8)}{2(p-2)(q-2)}\end{align}$$ From $p\not=q$ and $(1)$, one has $$1=-x-\frac{3}{2}\Rightarrow x=-\frac 52.$$
