Question

# Find the equation of the parabola if the focus is at (- 6, - 6) and the vertex is at (- 2,2).

Hint: Let’s find the intersection point of the axis and directrix, then find out the equation of the directrix then apply parabola property we will get the answer.

Let Z $({x_1},{y_1})$ be the coordinates of the point of intersection of the axis and the directrix of the parabola.

Then the vertex V (- 2,2) is the mid point of the line segment joining Z $({x_1},{y_1})$ and the focus S (- 6,- 6).

$\Rightarrow \dfrac{{{x_1} - 6}}{2} = - 2$ $\Rightarrow {x_1} = 2$

& $\dfrac{{{y_1} - 6}}{2} = 2 \Rightarrow {y_1} = 10$

Thus the directrix meets the axis at Z (2,10).

Let ${m_1}$ be the slope of the axis. Then,}

${m_1}$ = (Slope of the line joining the focus S and vertex V) = $\dfrac{{ - 6 - 2}}{{ - 6 + 2}} = \dfrac{{ - 8}}{{ - 4}} = 2$

$\Rightarrow$ Slope of the directrix which is perpendicular to axis is,

m = $- \dfrac{1}{m_1} = - \dfrac{1}{2}$

$\Rightarrow$ equation of directrix which is passing from (2,10) is

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

$\Rightarrow 2y + x - 22 = 0$

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

Distance of P from the focus = Perpendicular distance of P from the Directrix. (Parabola property)

$\Rightarrow \sqrt {{{\left( {x + 6} \right)}^2} + {{\left( {y + 6} \right)}^2}} = \left| {\dfrac{{2y + x - 22}}{{\sqrt {{2^2} + {1^2}} }}} \right|$

$\Rightarrow {\left( {x + 6} \right)^2} + {\left( {y + 6} \right)^2} = \dfrac{{{{\left( {2y + x - 22} \right)}^2}}}{5}$

$\Rightarrow 5{x^2} + 5{y^2} + 60x + 60y + 360 = 4{y^2} + {x^2} + 484 + 4xy - 44x - 88y$

$\Rightarrow 4{x^2} + {y^2} - 4xy + 104x + 148y - 124 = 0$

$\Rightarrow {\left( {2x - y} \right)^2} + 4\left( {26x + 37y - 31} \right) = 0$

So, this is your required equation of parabola.

NOTE: If two lines are perpendicular to each other, then the product of their slopes is -1.