# What is the equation of the parabola with a focus at (-3,1) and a directrix of y= 0?

May 9, 2018

The equation of parabola is $y = \frac{1}{2} {\left(x + 3\right)}^{2} + 0.5$

#### Explanation:

Focus is at $\left(- 3 , 1\right)$and directrix is $y = 0$. Vertex is at midway

between focus and directrix. Therefore vertex is at $\left(- 3 , \frac{1 - 0}{2}\right)$

or at $\left(- 3 , 0.5\right)$ . The vertex form of equation of parabola is

y=a(x-h)^2+k ; (h.k) ; being vertex. $h = - 3 \mathmr{and} k = 0.5$

Therefore vertex is at $\left(- 3 , 0.5\right)$ and the equation of parabola is

$y = a {\left(x + 3\right)}^{2} + 0.5$. Distance of vertex from directrix is

$d = 0.5 - 0 = 0.5$, we know $d = \frac{1}{4 | a |} \therefore 0.5 = \frac{1}{4 | a |}$ or

$| a | = \frac{1}{4 \cdot 0.5} = \frac{1}{2}$. Here the directrix is below

the vertex , so parabola opens upward and $a$ is positive.

$\therefore a = \frac{1}{2}$ . The equation of parabola is $y = \frac{1}{2} {\left(x + 3\right)}^{2} + 0.5$

graph{1/2(x+3)^2+0.5 [-10, 10, -5, 5]} [Ans]