# conic sections, parabola

• May 27th 2008, 02:56 PM
Tim28
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.
• May 27th 2008, 11:00 PM
earboth
Quote:

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:

$x^2-20x+y^2+4y+2xy=-28$
• May 28th 2008, 06:36 PM
Tim28
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.
• May 28th 2008, 09:19 PM
TheEmptySet
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.