Questions: Find an equation, in terms of x and y, for the parabola conic whose graph is shown.

Solution Steps

The vertex is the point on the parabola that is closest to the directrix. From the graph, the vertex is at $(-7, 0)$.

Since the parabola opens to the right and the directrix is vertical, the equation of the parabola is of the form $(y-k)^2 = 4p(x-h)$, where $(h,k)$ is the vertex. The vertex is $(-7, 0)$, so $h=-7$ and $k=0$. The directrix is $x = -11$. Since the parabola opens to the right, $p > 0$. The distance between the vertex and the directrix is $|h - (-11)| = |-7 + 11| = 4$. So, $p = 4$.

Substitute the values of $h$, $k$, and $p$ into the equation $(y-k)^2 = 4p(x-h)$: $(y-0)^2 = 4(4)(x-(-7))$ $y^2 = 16(x+7)$

Final Answer