Equation of parabola using axis and tangent at vertex

Question

Let $L=ax+by+c=0$ be the axis of a parabola, and the tangent at vertex be $M=bx-ay+d=0$

If $L^2=4pM$ is the equation of that parabola whose length of latus rectum is $4p$, then $a^2+b^2$ must be equal to $1$. But how does this condition come?

Answer 1

To understand this condition, we must understand a particular definition of the parabola.

Using your notation, let 2 perpendicular lines be

$ L = ax+by+c=0$

$M = bx-ay+d=0$

Then the locus of the point $(h,k)$ such that the square of its perpendicular distance from $L$ is proportional to its perpendicular distance from $M$ with a proportionality constant which is the length of the latus rectum, say $4p$.

Mathematically, it can be represented as follows:-

$$ \left( \frac {ah+bk+c}{\sqrt {a^2 +b^2}} \right)^2 = 4p \left( \frac {bh-ak+d}{\sqrt {a^2 +b^2}} \right) $$

Which upon on substitution and cancellation yields

$$ \frac{L^2}{\sqrt {a^2 +b^2}}= 4pM $$

This works for any parabola. Now applying your condition and comparing the equations, we can conclude that $\sqrt {a^2 +b^2} = 1$ and $ a^2 +b^2 = 1$