# How do you find the equation of a parabola with vertex at the origin and focus (0, -3/2)?

Mar 15, 2018

The equation of parabola is $y = - \frac{1}{6} {x}^{2}$

#### Explanation:

Focus is at $\left(0 , - \frac{3}{2}\right)$and vertex is at $0 , 0$. Vertex is at midway

between focus and directrix. Therefore ,directrix is $y = \frac{3}{2}$.

The vertex form of equation of parabola is

y=a(x-h)^2+k ; (h.k) ; being vertex. $h = 0 \mathmr{and} k = 0$

So the equation of parabola is $y = a {x}^{2}$

Here directrix is above the vertex , so parabola opens downward

and $a$ is negative. Distance of vertex from directrix is

$d = \frac{3}{2} - 0 = \frac{3}{2}$, we know $d = \frac{1}{4 | a |} \therefore \frac{3}{2} = \frac{1}{4 | a |}$ or

$| a | = \frac{2}{3 \cdot 4} = \frac{1}{6} \therefore a = - \frac{1}{6}$.

So the equation of parabola is $y = - \frac{1}{6} {x}^{2}$

graph{-1/6 x^2 [-10, 10, -5, 5]} [Ans]