Chapter 8.1, Problem 27E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions vertex $\left(2,-1\right)$ opens upward and focal width as $16$ .

The equation in the standard form for the parabola that satisfies conditions vertex $\left(2,-1\right)$ opens upward and focal width as $16$ is ${\left(x-2\right)}^2=16\left(y+1\right)$ .

Given information :

The vertex $\left(2,-1\right)$ opens upward and focal width as $16$ .

Formula used :

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

Calculation :

Consider the vertex $\left(2,-1\right)$ opens upward and focal width as $16$ ,

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

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

Equate $\left|4p\right|$ to $16$ .

  $\begin{array}{c}\left|4p\right|=16\\ \left|p\right|=4\end{array}$

This gives $p$ as either $4$ or $-4$ .

As $p>0$ , the value of $p$ is $4$ .

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

  $\begin{array}{c}{\left(x-2\right)}^2=4\left(4\right)\left(y-\left(-1\right)\right)\\ {\left(x-2\right)}^2=16\left(y+1\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions vertex $\left(2,-1\right)$ opens upward and focal width as $16$ is ${\left(x-2\right)}^2=16\left(y+1\right)$ .