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.

  • 2
    $\begingroup$ 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 $$(ax+by+c)^2=4p(bx−ay+d).$$ $\endgroup$ – Jan-Magnus Økland Mar 5 at 10:27
  • $\begingroup$ Oh I see what I was missing there... But still don't understand why we need that condition a^2+b^2=1. $\endgroup$ – AspiringEngineer Mar 5 at 13:07
  • $\begingroup$ It comes from $y^2=4px$ (orthonormally) rotated and translated to a general parabola. $\endgroup$ – Jan-Magnus Økland Mar 5 at 13:11
  • $\begingroup$ 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. $\endgroup$ – AspiringEngineer Mar 5 at 13:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Jan-Magnus Økland Mar 5 at 13:37

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