Chapter 8.1, Problem 19E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions vertex $\left(0,0\right)$ opens to downward and focal width as $6$ .

The equation in the standard form for the parabola that satisfies conditions vertex $\left(0,0\right)$ opens to downward and focal width as $6$ is $x^2=-6y$ .

Given information :

The parabola with vertex $\left(0,0\right)$ opens to downward and focal width as $6$ .

Formula used :

The equation for parabola for which focus $\left(0,p\right)$ is downward to the vertex is $x^2=4py$ .

Calculation :

Consider the parabola with vertex $\left(0,0\right)$ opens to downward and focal width as $6$ ,

As parabola opens to the downward, then $p<0$ .

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

Equate $\left|4p\right|$ to $6$ .

  $\begin{array}{c}\left|4p\right|=6\\ \left|p\right|=\frac{3}{2}\end{array}$

This gives $p$ as either $\frac{3}{2}$ or $-\frac{3}{2}$ .

As $p<0$ , the value of $p$ is $\frac{3}{2}$ .

Substitute this value into the equation $x^2=4py$ .

  $\begin{array}{c}{x}^2=4\left(-\frac{3}{2}\right)y\\ x^2=-6y\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions vertex $\left(0,0\right)$ opens to downward and focal width as $6$ is $x^2=-6y$ .