Question

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

Answer 1

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. 

Complete Step by step 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.