Question

The graph of a parabola is given. Match each graph to its equation.

$$\begin{gathered} A)y^2=4xC)y^2=-4xE){(y-1)}^2=4(x-1)G){(y-1)}^2=-4(x-1) \\ B)x^2=4yD)x^2=-4yF){(x+1)}^2=4(y+1)H){(x+1)}^2=4(y+1) \end{gathered}$$

Short answer

The correct match is (B).

Step-by-step solution

Step 1. Given information.

We have given graph is,

and given equations are,

$$\begin{gathered} A)y^2=4xC)y^2=-4xE){(y-1)}^2=4(x-1)G){(y-1)}^2=-4(x-1) \\ B)x^2=4yD)x^2=-4yF){(x+1)}^2=4(y+1)H){(x+1)}^2=4(y+1) \end{gathered}$$

Step 2. Concept used.

Standard equation of the parabola is:

${(y-k)}^2=4a(x-h)and{(x-h)}^2=4a(y-k)$

Where (h, k) is the vertex of the parabola and The focus is at (h, k + p). The directrix is the line $y=k-p$.

Step 3. Explanation.

We have given graph of the parabola is,

This parabola is upward parabola.

Vertex of the parabola is (0, 0) and point on the graph is (2, 1).

Substituting it in to the standard form of the parabola we get,

$$\begin{gathered} {(x-0)}^2=4a(y-0) \\ \Rightarrow x^2=4ay \\ \Rightarrow 2^2=4a\left(1\right) \\ \Rightarrow a=1 \\ Thatis,x^2=4y \end{gathered}$$

Step 4. Conclusion.

Hence, equation of the graph

is$x^2=4y$.