Question

How do you sketch the parabola `(y+3/2)^2=-7(x+9/2)` and find the vertex, focus, and directrix?

Answer 1

The vertex form of a parabola of this type is:
`x = 1/(4f)(y-k)^2+h`
where the vertex is `(h,k)`, the focus is `(h+f,k)`, and the equation of the directrix is `x = h-f`

Explanation:

Given:

`(y+3/2)^2=-7(x+9/2)`

Divide both sides of the equation by -7:

`1/-7(y+3/2)^2=(x+9/2)`

Please observe that `4f = -7` or `f = -7/4`

`1/(4(-7/4))(y+3/2)^2=(x+9/2)`

To comply with the form, there must be a minus sign inside the square:

`1/(4(-7/4))(y-(-3/2))^2=(x+9/2)`

Subtract `9/2` from both sides:

`x = 1/(4(-7/4))(y-(-3/2))^2-9/2`

Now, we may obtain the vertex, the focus, and the equation of the directrix by observation:

vertex `(-9/2,-3/2)`
focus`(-9/2-7/4,-3/2) = (-25/4,-3/2)`
The equation of the directrix `x = -9/2+7/4=-11/4`

Here is a graph of the equation:

graph{(y+3/2)^2=-7(x+9/2) [-10, 10, -5, 5]}