Question

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

and equations are,

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

Short answer

The correct match is (G).

Step-by-step solution

Step 1. Given information.

We have given graph is,

And equations of the parabola 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,

${\left(x-h\right)}^2=4a\left(y-k\right)\hspace{0.17em}and\hspace{0.17em}{\left(y-k\right)}^2=4a\left(x-h\right)$

Where (h, k) is the vertex of the parabola.

Step 3. Explanation.

We have given graph is,

This parabola is opens to the left and vertex of the parabola is, (1, 1).

Substituting it in to the standard formula of the hyperbola we will get,

${(y-1)}^2=4a(x-1)$

Using the point (0, 3),

$$\begin{gathered} {(3-1)}^2=4a(0-1) \\ \Rightarrow 4=-4a \\ \Rightarrow a=-1 \end{gathered}$$

Therefore equation of the parabola is,

${(y-1)}^2=-4(x-1)$

Step 4. Conclusion.

Hence, equation of the parabola for the graph

is${(y-1)}^2=-4(x-1)$.