Find the equation of the parabola if

the focus is at (−6, −6) and the vertex is at (−2, 2)

#### Solution

In a parabola, the vertex is the mid-point of the focus and the point of intersection of the axis and the directrix.

Let (*x _{1}, y_{1}*) be the coordinates of the point of intersection of the axis and directrix.

It is given that the vertex and the focus of a parabola are (−2, 2) and (−6, −6), respectively.

∴ Slope of the axis of the parabola =\[\frac{- 6 - 2}{- 6 + 2} = \frac{- 8}{- 4} = 2\]

Slope of the directrix =\[\frac{- 1}{2}\]

Let the directrix intersect the axis at *K *(*r*, *s*).

∴\[\frac{r - 6}{2} = - 2, \frac{s - 6}{2} = 2\]

∴ Required equation of the directrix:

\[y - 10 = \frac{- 1}{2}\left( x - 2 \right)\]

⇒ \[2y + x - 22 = 0\]

Now, let *P* (*x*, *y*) be any point on the parabola whose focus is *S* (−6, −6), and the directrix is \[2y + x - 22 = 0\]

Draw *PM *perpendicular to \[2x + y + 22 = 0\]

Then, we have:

\[SP = PM\]

\[ \Rightarrow S P^2 = P M^2 \]

\[ \Rightarrow \left( x + 6 \right)^2 + \left( y + 6 \right)^2 = \left( \frac{2y + x - 22}{\sqrt{5}} \right)^2 \]

\[ \Rightarrow 5\left( x^2 + 12x + 36 + y^2 + 12y + 36 \right) = 4 y^2 + x^2 + 484 + 4xy - 88y - 44x\]

\[ \Rightarrow 4 x^2 + y^2 - 4xy + 104x + 148y - 124 = 0\]

\[ \Rightarrow \left( 2x - y \right)^2 - 4\left( 26x + 37y - 31 \right) = 0\]