What is the equation of the parabola with a focus at (3,-8) and a directrix of y= -5?

Jun 26, 2018

The equation is $y = - \frac{1}{6} {\left(x - 3\right)}^{2} - \frac{39}{6}$

Explanation:

Any point $\left(x , y\right)$ on the parabola is equidistant from the directrix and from the focus.

Therefore,

$\left(y + 5\right) = \sqrt{{\left(x - 3\right)}^{2} + {\left(y + 8\right)}^{2}}$

Squaring both sides

${\left(y + 5\right)}^{2} = {\left(x - 3\right)}^{2} + {\left(y + 8\right)}^{2}$

${y}^{2} + 10 y + 25 = {\left(x - 3\right)}^{2} + {y}^{2} + 16 y + 64$

$6 y = - {\left(x - 3\right)}^{2} - 39$

$y = - \frac{1}{6} {\left(x - 3\right)}^{2} - \frac{39}{6}$

graph{(y+1/6(x-3)^2+39/6)(y+5)=0 [-28.86, 28.87, -14.43, 14.45]}

Jun 26, 2018

The equation of parabola is $y = - \frac{1}{6} {\left(x - 3\right)}^{2} - 6.5$

Explanation:

Focus is at $\left(3 , - 8\right)$and directrix is $y = - 5$. Vertex is at midway

between focus and directrix. Therefore,vertex is at $\left(3 , \frac{- 5 - 8}{2}\right)$

or at $\left(3 , - 6.5\right)$ . The vertex form of equation of parabola is

y=a(x-h)^2+k ; (h,k)  being vertex. $h = 3 \mathmr{and} k = - 6.5$

So the equation of parabola is $y = a {\left(x - 3\right)}^{2} - 6.5$. Distance of

vertex from directrix is $d = | 6.5 - 5 | = 1.5$, we know $d = \frac{1}{4 | a |}$

$\therefore 1.5 = \frac{1}{4 | a |} \mathmr{and} | a | = \frac{1}{1.5 \cdot 4} = \frac{1}{6}$. Here the directrix is above

the vertex , so parabola opens downward and $a$ is negative.

$\therefore a = - \frac{1}{6}$. Hence the equation of parabola is

$y = - \frac{1}{6} {\left(x - 3\right)}^{2} - 6.5$

graph{-1/6(x-3)^2-6.5 [-40, 40, -20, 20]}