1. ## conic sections, parabola

What if the directrix is not parallel to an axis? Suppose it is y=x+2, and let the focus be ((4,0)
How do I derive the equation of the parabola? The distance from the parabola is perpendicular to the directrix.

2. Originally Posted by Tim28
What if the directrix is not parallel to an axis? Suppose it is y=x+2, and let the focus be ((4,0)
How do I derive the equation of the parabola? The distance from the parabola is perpendicular to the directrix.
1. Use the complete definition of the parabola:
The distance of an arbitrary point P of the parabola to the focus F is equal to the perpendicular distance to the directrix d (= blue lines)

2. The triangle QFP is isosceles and the line MP is the perpendicular bisector of QF (by the way: MP is simultaneously the tangent to the parabola)

3. As posted the directrix has the equation y = x+2 and the focus is F(4,0). The point Q is located on the directrix and has the coordinates Q(q, q+2)

4. Then M has the coordinates $M \left( \frac12q+2~,~ \frac12 q + 1\right)$

5. The slope of QF is $m_{QF}=\frac{q+2}{q-4}$ and therefore the perpendicular slope is $m = -\frac{q-4}{q+2}$

6. The equation of the perpendicular bisector of QF is:

$y-\left(\frac12 q + 1\right)=-\frac{q-4}{q+2} \left(x-\left(\frac12q+2\right) \right)$ which will yield: $\boxed{ y= -\frac{q-4}{q+2} x + \frac{q^2+2q-6}{q+2}}$

7. The perpendicular line to directrix passing through Q has the equation:

$y-(q+2)=-(x-q)~\implies~\boxed{y=-x+2q+2}$

8. Now calculate the coordinates of the point P which is the intersection between the lines of #6 and #7:

$-\frac{q-4}{q+2} x + \frac{q^2+2q-6}{q+2} = -x+2q+2$

After a lot of transformation you'll get:

$x = \frac16\cdot (q^2+4q+10)$

9. Plug in this value into the equation at #7 to calculate the y-coordinate of P:

$y = -\frac16 \cdot (q^2-8q-2)$

10. Now you have a parametric equation of the parabola:

$\left|\begin{array}{l} x = \frac16\cdot (q^2+4q+10) \\ y = -\frac16 \cdot (q^2-8q-2) \end{array} \right.$

11. Solve $x = \frac16\cdot (q^2+4q+10)$ for q and plug this term into $y = -\frac16 \cdot (q^2-8q-2)$ which will yield:

$x^2-20x+y^2+4y+2xy=-28$

3. ## Thank you very much

I cannot imagine a more complete, more lucid analysis. I have copied your reply and will be studying it in detail. My only regret is that I couldn't do it myself.

4. ## Another solution

After Jen and I saw Earboths amazing post, she said to me can't you do that with a matrix rotation. I thought yes, you should be able to. So Here it is.

The directrix $y=x+2$ can be written as a vector

$v_1=(t,t+2)$ and the focus as $f=(4,0)$.

Since the directrix has slope 1 I will rotate it 45 degrees clockwise.

$R=\begin{bmatrix} \cos \left(-\frac{\pi}{4}\right) && -\sin\left(-\frac{\pi}{4}\right) \\ \sin \left( -\frac{\pi}{4}\right) && \cos \left( -\frac{\pi}{4}\right)\end{bmatrix}=\begin{bmatrix} \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix}$

Note that the inverse of R is $R^T$ the transpose of R.

Now we can transform the focus and the directrix

$v_{1R}=\begin{bmatrix} \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2} \\ -\frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix} \begin{bmatrix} t \\ t+2 \end{bmatrix}=\begin{bmatrix} (t+1)\sqrt{2} \\ \sqrt{2} \end{bmatrix}$

$f_R=(2\sqrt{2},-2\sqrt{2})$

Now in the rotated system the equation of the directrix is
$y_R=\sqrt{2}$

Now the form of the parabola is $(x-h)^2=4p(y-k)$

We can identify h and k as $h=2\sqrt{2}\\\ k=\frac{\sqrt{2}-2\sqrt{2}}{2}=-\frac{\sqrt{2}}{2}$
p is the vertical distance from the vertex to the focus
$p=-2\sqrt{2}+\frac{\sqrt{2}}{2}=-\frac{3}{2}\sqrt{2}$

So we get the equation

$(x-2\sqrt{2})^2=-6\sqrt{2}(y+\frac{\sqrt{2}}{2})$

Solving this for y gives

$y=-\frac{\sqrt{2}}{12}x^2-\frac{2}{3}x+\frac{7}{6}\sqrt{2}$

Now we can write this in vector form

$v_{pR}=\begin{bmatrix} s \\ -\frac{\sqrt{2}}{12}s^2-\frac{2}{3}s+\frac{7}{6}\sqrt{2}\end{bmatrix}$

Now to rotate back we use $R^T$

$R^Tv_{pR}=\begin{bmatrix} \frac{\sqrt{2}}{2} && -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} && \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} s \\ -\frac{\sqrt{2}}{12}s^2-\frac{2}{3}s+\frac{7}{6}\sqrt{2}\end{bmatrix}=\beg in{bmatrix} \frac{1}{12}s^2+\frac{1}{6}\sqrt{2}s+\frac{7}{6} \\ -\frac{1}{12}s^2+\frac{5}{6}\sqrt{2}s-\frac{7}{6} \end{bmatrix}$

Now we have parametric equation for the parabola

$x= \frac{1}{12}s^2+\frac{1}{6}\sqrt{2}s+\frac{7}{6}$

$y=-\frac{1}{12}s^2+\frac{5}{6}\sqrt{2}s-\frac{7}{6}$

Now we can eliminate the parameter

$x+y=s\sqrt{2} \iff \frac{(x+y)^2}{2}=s^2$

$5x-y=\frac{1}{2}s^2+7 \iff s^2=10x-2y-14$

So we end up with $\frac{(x+y)^2}{2}=10x-2y-14$

This turns out to be the same as Earboth's if you move some stuff around.