Question

Find the equation of the parabola if 

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

Answer 1

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

Let (x₁, y₁) 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). 

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

\[ \Rightarrow r = 2, s = 10\]

∴ 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:

\[SP = PM\]
\[ \Rightarrow S P^2 = P M^2 \]
\[ \Rightarrow \left( x + 6 \right)^2 + \left( y + 6 \right)^2 = \left( \frac{2y + x - 22}{\sqrt{5}} \right)^2 \]
\[ \Rightarrow 5\left( x^2 + 12x + 36 + y^2 + 12y + 36 \right) = 4 y^2 + x^2 + 484 + 4xy - 88y - 44x\]
\[ \Rightarrow 4 x^2 + y^2 - 4xy + 104x + 148y - 124 = 0\]
\[ \Rightarrow \left( 2x - y \right)^2 - 4\left( 26x + 37y - 31 \right) = 0\]