How can I find the equation of a parabola with focus and directrix?

Question

Focus: (3,5)

Directrix: $y=-x-1$

I'm comfortable with questions where the directrix is simply $y=3$ or something that is constant (i.e. a horizontal line).

I know that I need to find the distance between a general point (x,y) and (3,5) as well as the distance between (x,y) and the line $y=-x-1$

The distance between (x,y) and (3,5) is $\sqrt{(x-3)^2+(y-5)^2}$

I see online that the formula for the distance from a line to a point is given by

$\frac{|Ax+By+C|}{\sqrt{A^2+B^2}}$

I want to know if there is some easier way for me to calculate this distance? Do I just need to memorize this?

Also, using this formula, is my calculation correct?

Distance from $(x,y)$ to $y=-x-1$:

$$x+y+1=0$$

$$ \frac{|x+y+1|}{\sqrt{1^2+1^2}}$$

$$\sqrt{(x-3)^2+(y-5)^2} = \frac{|x+y+1|}{\sqrt{2}}$$

$$(x-3)^2+(y-5)^2=\frac{(x+y+1)^2}{2}$$

Then from here I just expand everything out and solve for $x$:

$$2(x^2-6x+9+y^2-10y+25)=x^2+y^2+2xy+2x+2y+1$$

$$2x^2+2y^2-12x-20y+68=x^2+y^2+2xy+2x+2y+1$$

$$x^2=-67+14x+22y+2xy-y^2$$

Not exactly sure what to do from here?

Answer 1

First, you can find the vertex of the parabola. It lies half way between the focus and the directrix.

The axis of the parabola passes through the focus, and is perpendicular to the directrix. The directrix is

$x + y + 1 = 0$

So the unit vector along the axis is

$u_1 = [1, 1]^T / \sqrt{2} $

The distance between the focus and the directrix is

$2 p = \dfrac{|3 + 5 + 1 | }{\sqrt{2} } = \dfrac{9}{\sqrt{2}}$

Therefore the vertex of the parabola is

$ V = (3, 5) - \dfrac{9}{2 \sqrt{2}} (1, 1)/ \sqrt{2} = (0.75, 2.75) $

$p = \dfrac{9}{2 \sqrt{2}} $ is the focal length.

Rotate $u_1$ by ($90^\circ$ clockwise) to generate the other axis

$ u_2 = (1, -1) / \sqrt{2} $

Now in the coordinate refernce frame specified by $u_1, u_2$ the equation of the parabola is

$ y' = \dfrac{1}{4 p} x'^2 $

where $x'$ is along $u_2$ and $y'$ is along $u_1$

Hence

$ y' = \dfrac{\sqrt{2}}{18} x'^2 $

Finally to get the equation in the $xy$ frame, we note that

$(x, y) = V + [u_2, u_1] (x', y') $

Thus

$ (x', y') = [u_2,u_1]^{-1} ( (x, y) - V ) $

Now:

$[u_2, u_1]^{-1} = \sqrt{2} \begin{bmatrix} 1 && 1\\-1&&1 \end{bmatrix}^{-1} = \dfrac{1}{\sqrt{2}} \begin{bmatrix} 1 && -1 \\ 1 && 1 \end{bmatrix} $

Writing $x', y'$ explicitly in terms of $x, y$, we have

$x' = \dfrac{1}{\sqrt{2}} (x - y + 2) $

and

$y' = \dfrac{1}{\sqrt{2}} ( x + y - 3.5 ) $

Plug these two expressions into

$ y' = \dfrac{\sqrt{2}}{18} x'^2 $

You get

$ 9 (x + y - 3.5) = \dfrac{1}{2} ( x- y + 2 )^2 $

Therefore,

$ 18 (x + y - 3.5) = x^2 + y^2 + 4 + 4 x - 4 y - 2 xy $

and finally,

$ x^2 - 2 xy + y^2 - 14 x - 22 y + 67 = 0 $

which is exactly what you got.