# General form of parabola using equation of Tangent at Vertex and Axis

I had read in my notes that the equation of parabola can be given by

(Equation of axis)$$^2$$ = (Length of Latus rectum)*(Equation of tangent at vertex)

(I don't know the systematic proof. Is there something I am missing in the equation?)

Now take look at this very basic equation of a parabola

$$y^2=4ax$$

Here the equation of axis of parabola is $$(y=0)$$ and that of tangent at vertex is $$(x=0)$$

I can also write the equation of axis as $$(ny=0)$$ and tangent at vertex as $$(mx=0)$$
(where m and n are constants)

And hence using the first equation I can write the equation of parabola as

$$(ny)^2 = 4a(mx)$$

which gives me a completely different parabola.
I don't know where I have gone wrong. Please guide me.

• When $a^2+b^2=1$ and the line $ax+by+c=0$ is the axis of a parabola then its tangent at the vertex is perpendicular to it, say $bx−ay+d=0$. And an equation of the parabola is $$(ax+by+c)^2=4p(bx−ay+d).$$ – Jan-Magnus Økland Mar 5 at 10:27
• Oh I see what I was missing there... But still don't understand why we need that condition a^2+b^2=1. – AspiringEngineer Mar 5 at 13:07
• It comes from $y^2=4px$ (orthonormally) rotated and translated to a general parabola. – Jan-Magnus Økland Mar 5 at 13:11
• Take a look at this parabola. (2x - y - 3)^2 = -20(x + 2y - 4). It doesn't satisfy the condition but represents a parabola with the given lines as its axis and tangent at vertex. – AspiringEngineer Mar 5 at 13:23
• You can rewrite it: $5(\frac{2x-y-3}{\sqrt{5}})^2=-20\sqrt{5}\frac{(x+2y-4)}{\sqrt{5}}$ i.e. $(\frac{2x-y-3}{\sqrt{5}})^2=-4\sqrt{5}\frac{(x+2y-4)}{\sqrt{5}}$ and identify $p$. By the way it is also $$5\,\left(-\left({{2\,y}\over{\sqrt{5}}}+{{x}\over{\sqrt{5}}}-{{9 }\over{\sqrt{5}}}\right)^2+\left(y+1\right)^2+\left(x-1\right)^2 \right)$$ making $(1,-1)$ the focus and $x+2y-9$ the directrix. – Jan-Magnus Økland Mar 5 at 13:37