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

1 Answer
Dec 18, 2017

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]}