Chapter 8.1, Problem 30E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions vertex $\left(2,3\right)$ opens right and focal width as $5$ .

The equation in the standard form for the parabola that satisfies conditions vertex $\left(2,3\right)$ opens right and focal width as $5$ is ${\left(y-3\right)}^2=5\left(x-2\right)$ .

Given information :

The parabola with vertex $\left(2,3\right)$ opens right and focal width as $5$ .

Formula used :

The equation for parabola for which vertex is at $\left(h,k\right)$ and opens right is ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

Calculation :

Consider the parabola with vertex $\left(2,3\right)$ opens right and focal width as $5$ ,

As parabola opens to the right, then $p>0$ .

The width of focal will be $\left|4p\right|$ .

Equate $\left|4p\right|$ to $5$ .

  $\begin{array}{c}\left|4p\right|=5\\ \left|p\right|=\frac{5}{4}\end{array}$

This gives $p$ as either $\frac{5}{4}$ or $-\frac{5}{4}$ .

As $p>0$ , the value of $p$ is $\frac{5}{4}$ .

Substitute the desired values to the equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

  $\begin{array}{c}{\left(y-3\right)}^2=4\left(\frac{5}{4}\right)\left(x-2\right)\\ {\left(y-3\right)}^2=5\left(x-2\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions vertex $\left(2,3\right)$ opens right and focal width as $5$ is ${\left(y-3\right)}^2=5\left(x-2\right)$ .