Let's say we have a vertex of the parabola $V(2,6)$ and a point where it passes through $(-1, 4)$. It follows that $h=2$ and $k=6$ and the point where it passes through is a solution of the equation that is $x=-1$ and $y=4$. By substituting the value to the vertex form of the parabola equation we can easily get its equation. But when I try to use the conics form of the parabola equation, I became confused especially in the $4p$ part of the equation: $(x-h)^2=4p(y-k)$

When solving this type of problems do I need to use the vertex form of the parabola equation?


Usually it is $$(x-h)^2 = 2p (y-k)$$ and since $V(2,6)$ is vertex we have now: $$(x-2)^2= 2p(y-6)$$ Plugging $(-1,4)$ in to that equation we get: $$ 9 = -4p\implies p = -{9\over 4}$$

you can also swap $x,y$ and get another parabola:$$(y-h)^2 = 2p (x-k)$$ so $$(y-6)^2 = 2p (x-2)$$ so $4= -6p$ so $p =-{2\over 3}$.

  • $\begingroup$ Does that mean that I can have two parabolas containing those points? $\endgroup$ – sammyyy Aug 7 '18 at 14:34
  • $\begingroup$ Ok. I've got p= -9/4 at first when I used the conics form. I thought it as wrong as I get different equation using vertex form. Thank you for clarifying $\endgroup$ – sammyyy Aug 7 '18 at 14:54

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