The question is as follows:

Consider the parabola whose focus is $F = (1, 4)$ and whose directrix is the line $x = -3$. Sketch the parabola, and make calculations that confirm that $P = (7,12)$ is on it.

I found the equation of the line to be $\frac{-8x + 8 - y^2}{8} = y$, however, I don't think that it is correct. I have been trying to trace my error in my algebra. Here's the equation that I used to find the parabola's equation.

$$\sqrt{(1-x)^2 + (4-y)^2} = \sqrt{(-3-x)^2}$$ $$x^2 + 2x + 1 + y^2 - 8y + 16 = x^2 + 6x + 9$$ $$-2x + 17 + y^2 -8y= 6x + 9$$ $$\text{Which then finally led me to}: \frac{-8x + 8 - y^2}{8} = y$$

Are any of my steps incorrect, and, if so, where?

  • 2
    $\begingroup$ The second equation is wrong, $(1-x)^2 = x^2 - 2x + 1$ not $x^2 + 2x + 1$. Also use the fact that the directrix is the same distance away from the vertex as the focus point is from the vertex. That should tell you where the vertex of your parabola is. Then, using this and the fact that $x = -3$ write the equation of your parabola as $(x-h)^2 = 4p (y-k)$ where $p$ is the distance from the vertex to the focus point and $(h,k)$ is the position of your vertex. With that equation you should get all the info you need. $\endgroup$ – R.Mor Sep 17 '17 at 16:48
  • $\begingroup$ He perhaps made a typo in the second line, but the mistake does not carry through the next line. $\endgroup$ – paulinho Sep 17 '17 at 18:31
  • $\begingroup$ I see, then everything he made is correct. $\endgroup$ – R.Mor Sep 17 '17 at 19:04

Since the directrix is vertical, the parabola is horizontal, so you should solve in terms of $x$. There is a mistake in your last line - it should be $+y^2$ in the numerator. However, solving in terms of $x$, we get $$x=\frac{y^2-8y+8}{8}$$ and you will see that the point $(7,12)$ is on the parabola.


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