Chapter 9, Problem 8CR

To determine

To find vertex of the parabola $4x=y^2-4y$ .

Answer to Problem 8CR

The Vertex is $b.$

  $\left(-1,2\right)$ .

Explanation of Solution

Given:

Equation of parabola: $4x=y^2-4y$

Formula Used:

Parabola equation is given as ${\left(y-k\right)}^2=4p\left(x-h\right)$ ; where $\left(h,k\right)$ is the Vertex and Directrix is $x=h-p$ .

Calculation:

Given:

Equation of parabola: $4x=y^2-4y$

  $\begin{array}{l}{y}^2-4y=4x\\ y^2-4y+4-4=4x\\ {\left(y-2\right)}^2=4x+4\\ {\left(y-2\right)}^2=4\left(x+1\right)\end{array}$

Using the above formula and comparing the above values in the equation of the Parabola, we have:

  $k=2$

And $h=-1$

Thus,

Vertex $=$$\left(h,k\right)$

Vertex $=$$\left(-1,2\right)$

Thus, the Vertex is $\left(-1,2\right)$ .

Conclusion:

The Vertex is $b.$

  $\left(-1,2\right)$ .