# How do you find the equation of the parabola having its focus at ( 1/2, 1/2) and the directrix along x - y = 1?

Oct 7, 2016

${x}^{2} - 4 x - 4 y - 2 x y + {y}^{2} = 0$

#### Explanation:

Calling the directrix as

${L}_{d} \to p = {p}_{0} + {\lambda}_{d} {\vec{v}}_{d}$ with
${p}_{0} = \left(0 , - 1\right)$, ${\vec{v}}_{d} = \left(1 , - 1\right)$ and ${\lambda}_{d} \in \mathbb{R}$

and the symmetry line

${L}_{s} \to p = {p}_{f} + {\lambda}_{s} {\vec{v}}_{s}$ with
${p}_{f} = \left(\frac{1}{2} , \frac{1}{2}\right)$, ${\vec{v}}_{s} = \left(1 , 1\right)$ and ${\lambda}_{s} \in \mathbb{R}$

we have $O = {L}_{d} \cap {L}_{s}$ computed as

${p}_{0} + {\lambda}_{d} {\vec{v}}_{d} = {p}_{f} + {\lambda}_{s} {\vec{v}}_{s}$.

Solving for ${\lambda}_{d} , {\lambda}_{s}$ we obtain ${\lambda}_{d} = - \frac{1}{2} , {\lambda}_{s} = - 1$ and consequently
$O = \left(- \frac{1}{2} , - \frac{1}{2}\right)$

Now, the parabola is the place where the distance between a generic plane point $q = \left(x , y\right)$ and ${L}_{d}$ is equal to $\left\lVert q - {p}_{f} \right\rVert$.

but

$\min \left\lVert q - {L}_{d} \right\rVert = \sqrt{{\left\lVert q - O \right\rVert}^{2} - {\left\langleq - O , \hat{{v}_{d}}\right\rangle}^{2}}$

where $\hat{{v}_{d}} = {\vec{v}}_{d} / \left\lVert {\vec{v}}_{d} \right\rVert$ so the parabola is the place of points $q$ such that

${\left\lVert q - O \right\rVert}^{2} - {\left\langleq - O , \hat{{v}_{d}}\right\rangle}^{2} = {\left\lVert q - {p}_{f} \right\rVert}^{2}$

giving

${x}^{2} - 4 x - 4 y - 2 x y + {y}^{2} = 0$

Attached the parabola plot Oct 8, 2016

${x}^{2} + {y}^{2} + x y - 4 y = 0$, with vertex at V(3/4, 1/4) and size $a = \frac{\sqrt{5}}{4}$.

#### Explanation:

Let (x, y) be a point on the parabola.

Using that its distance from the focus equals the distance from the

directrix,

sqrt((x-1/2)^2+(y-1/2)^2))=+-(x-y-1)/sqrt(1^2+(-1)^2), two halves of the

parabola. Squaring,

${x}^{2} + {y}^{2} - x - y + \frac{1}{2} = \left(\frac{1}{2}\right) \left({x}^{2} + {y}^{2} + 1 - 2 x y - 2 x + 2 y\right)$. Upon simplification,

${x}^{2} + {y}^{2} + x y - 4 y = 0$

The vertex V of the parabola is on the perpendicular SX from the

focus S(1/2, 1/2), on the directrix DX given by x-y=1. The equation of

the perpendicular is of the form x+y=c. As S(1/2, 1/2) lies on this, c = 1.

$x - y = 1 \mathmr{and} x + y = 1$ meet at X( 1, 0).

V bisects SX.

So, V is (3/4, 1/4).. The size of the parabola

a = VS = $\sqrt{{\left(\frac{3}{4} - \frac{1}{2}\right)}^{2} + {\left(\frac{1}{4} - \frac{1}{2}\right)}^{2}} = \sqrt{\frac{1}{16} + \frac{1}{4}} = \frac{\sqrt{5}}{4}$