Chapter 9.3, Problem 3OE

To determine

To find: The vertex of the parabola.

Answer to Problem 3OE

The required vertex is $V\left(2,6\right)$ .

Explanation of Solution

Given information: It is given that $F\left(4,6\right)$ and $D:x=0$ .

Formula used: For parabola: $y-k=\frac{1}{4c}{\left(x-h\right)}^2$ :

  $\begin{array}{l}Vertex\left(h,k\right)\\ Focus\left(h+c,k\right)\\ D:x=k-c\end{array}$

Calculation: Since $F\left(4,6\right)$

  $\begin{array}{l}\because F\left(h+c,k\right)=\left(4,6\right)\\ \therefore h+c=4\&k=6\end{array}$

Now, $D:x=0$

  $\begin{array}{l}x=h-c\\ 0=h-c\\ \therefore h=c\end{array}$

Now, put $h=c$ in $h+c=4$

  $\begin{array}{l}h+h=4\\ 2h=4\\ h=2\end{array}$

Now, put the value of $h\&k$ ,

  $\begin{array}{l}\therefore \left(h,k\right)=\left(2,6\right)\\ \therefore V\left(2,6\right)\end{array}$

Therefore, the required vertex is $V\left(2,6\right)$ .