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Find the equation of the paraboles, with:

  • Focus $(3,0)$ and $x=-3$ is the directrix

  • Focus $(0,2)$ and $y=-2$ is the directrix

  • Vertex (I believe it is the vertex, the lowest/highest point) $(1,2)$ and $x=-1$ is the directrix

  • What is the focus and directrix of the parabola $(y-2)^2 = 4(x-4)$

I don't know how to do these, I only know this formula:

$ y - b = \dfrac{1}{4c}x^2$ which is a parabola with the vertex $T(a,b)$ and locus $F(a,b+c)$

However, I don't understand what the standard formula becomes when the directrix is a vertical line instead of a horizontal line. I don't think I grasp parabolas too well.. I would like some help with these questions (a push in the right direction) and maybe a link.

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By definition, parabola is the locus of the points equidistant from a given point (focus) and a given straight line (directrix).

Choose the coordinate frame such that the directrix is parallel to the $Y$ axis running at the distance of $\frac{p}{2}$ below it, that is $x_d=-\frac{p}{2}$, and the focus be at $F(\frac{p}{2},0)$. Then for an arbitrary point $M(x,y)$ on the parabola: $$\left|x+\frac{p}{2}\right|=\sqrt{(x-\frac{p}{2})^2+y^2}$$ Squaring and expanding the RHS we obtain, after some cancellation, the canonical equation of the parabola: $$y^2=2px$$ You should work through the calculation and draw the picture to fully understand the choice of the frame and the constant.

Now in your case the orientation of the parabola is clearly the same as in the canonical derivation, hence the shifts along the vertical axis are irrelevant. After shifting the frame along $X$ to the right by 4 we also see that $p=2$. The rest should follow easily.


  • Focus $(3,0)$ and $x=-3$ is the directrix

In this case $\frac{p}{2}=3$ so the equation is $y^2=12x$

  • Focus $(0,2)$ and $y=-2$ is the directrix

$\frac{p}{2}=2$ and the roles of coordinates are swapped, so $x^2=8y$

  • Vertex (I believe it is the vertex, the lowest/highest point) $(1,2)$ and $x=-1$ is the directrix

$\frac{p}{2}=1$, and we also need to make a vertical shift so $y\to y-2$, hence $(y-2)^2=2x$

  • What is the focus and directrix of the parabola $(y-2)^2 = 4(x-4)$

Move the origin to $(4,2)$ temporarily by changing coordinates $x_1=x-4$, $y_1=y-2$. In the new coordinates the equation is $y_1^2=4x_1$, so $p=2$. Directrix is given by equation $x_1=-1$ and focus is at $(1,0)_{(x_1,y_1)}$. Coming back to $x$ and $y$ we obtain $x=3$ and focus is at $(5,2)$

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So canonical means it's vertex is pointed leftward/rightward instead of upward/downward? And another (relevant) question. What does 'coincide with an axis' mean? Does it mean have a point of tangency with the axis? – JohnPhteven Nov 2 '12 at 21:32
I still don't understand this at all – JohnPhteven Nov 2 '12 at 21:40
This doesn't answer my questions, I knew the formula and the proof of it already.. – JohnPhteven Nov 2 '12 at 21:54
well, if you really follow the proof you realize that the solution amounts to just plugging constants into it – Valentin Nov 2 '12 at 22:10

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