Find an equation for the conic whose graph is shown.
A) $x ^ { 2 } = 4 ( y + 3 )$
B) $x ^ { 2 } = 3 ( y + 3 )$
C) $y ^ { 2 } = 12 ( x + 3 )$
D) $\frac { x } { 12 } = ( y + 3 ) ^ { 2 }$
E) $x ^ { 2 } = 12 ( y + 3 )$

Question

Find an equation for the conic whose graph is shown. 
A) $x ^ { 2 } = 4 ( y + 3 )$ 
B) $x ^ { 2 } = 3 ( y + 3 )$ 
C) $y ^ { 2 } = 12 ( x + 3 )$ 
D) $\frac { x } { 12 } = ( y + 3 ) ^ { 2 }$ 
E) $x ^ { 2 } = 12 ( y + 3 )$

Quizplus · Accepted Answer

The conic is clearly a parabola opening upwards, with its vertex at (-3,-3). Since the axis of symmetry is vertical, the equation must be of the form $x^2 = 4py$, where p is the distance from the vertex to the focus. Since the focus is 12 units above the vertex, we have $4p = 12$ or $p=3$. Substituting into the general equation gives $x^2 = 4(3)(y+3)$, which simplifies to $x^2 = 12(y+3)$. Therefore, the answer is E.

Find an Equation for the Conic Whose Graph Is Shown x2=4(y+3)x ^ { 2 } = 4 ( y + 3 )x2=4(y+3)

Find an Equation for the Conic Whose Graph Is Shown $x ^ { 2 } = 4 ( y + 3 )$