Find the equation for the parabola with focus at (3, -3) and vertex at (3, 1).
a. x^2 + 6x +16y -7 = 0
b. x^2 + 6x + 16y - 25 = 0
c. x^2 - 6x - 16y + 25 = 0
d. x^2 - 6x + 16y -7 = 0

Question

Find the equation for the parabola with focus at (3, -3) and vertex at (3, 1).  
a. x^2 + 6x +16y -7 = 0  
b. x^2 + 6x + 16y - 25 = 0  
c. x^2 - 6x - 16y + 25 = 0  
d. x^2 - 6x + 16y  -7 = 0

mertsj · Answer

It's d.  Do you care to see why or do you just want the answer?

anonymous · Answer

No, please explain.

mertsj · Answer

Since the vertex is (3,1) and the focus is (3,-3), the parabola must be concave downward since the focus in always "inside" the parabola.  That means the a value must be negative.  Also the vertex is (3,1) so if we put each parabola in vertex form we can identify the correct equation.  For Letter D it goes like this:

mertsj · Answer

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