Find the equation of a parabola with vertex at the origin and focus (0, - 32).

y^2 = - 128x

x^2 = - 32y

x^2 = 128y

x^2=-128y

Question

Mathematics

Jefferson Blackwell

19 January, 10:33

Find the equation of a parabola with vertex at the origin and focus (0, - 32).

y^2 = - 128x

x^2 = - 32y

x^2 = 128y

x^2=-128y

+4

Answers (1)

Know the Answer?

Stacy · Answer 1

x^2 = - 128y

Step-by-step explanation:

The general formula for this type of parabola is x^2 = 4ay where a is the y coordinate of the focus. So 4a = 4*-32 = 1-28.

So the answer is x^2 = - 128y

Find the equation of a parabola with vertex at the origin and focus (0, - 32).y^2 = - 128xx^2 = - 32yx^2 = 128yx^2=-128y

Find the equation of a parabola with vertex at the origin and focus (0, - 32).

y^2 = - 128x

x^2 = - 32y

x^2 = 128y

x^2=-128y