Question 11.2.2: Finding the Equation of a Parabola Find the standard equatio...

a. The vertex (0, 0) of the parabola and its focus (-3, 0) are on the x-axis; so the parabola opens to the left. Its equation must be of the form y^{2}=-4 a x \text { with } a=3.

\begin{array}{ll}y^{2}=-4 a x & \text { Form of the equation } \\y^{2}=-4(3) x & \text { Replace } a \text { with } 3 \\y^{2}=-12 x & \text { Simplify }\end{array}

The equation of the parabola is y^{2}=-12 x.

b. Because the vertex of the parabola is (0, 0), the axis is the y-axis, and the point (-4, 0) is above the x-axis, the parabola opens up and the equation of the parabola is of the form

x^{2}=4 a y

This equation is satisfied by x=-4 \text { and } y=2 because the parabola passes through the point (-4, 0).

\begin{aligned}(-4)^{2} &=4 a(2) & & \text { Replace } x \text { with }-4 \text { and } y \text { with } 2 \\16 &=8 a & & \text { Simplify. } \\2 &=a & & \text { Divide both sides by } 8\end{aligned}

Therefore, the equation of the parabola is

\begin{array}{ll}x^{2}=4(2) y & \text { Replace } a \text { with } 2 \text { in } x^{2}=4 a y \\x^{2}=8 y & \text { Simplify. }\end{array}

The required equation of the parabola is x^{2}=8 y.