Chapter 8.1, Problem 22E

Solution

To calculate : The equation in the standard form for the parabola that satisfies conditions focus $\left(-5,3\right)$ and vertex $\left(-5,6\right)$ .

The equation in the standard form for the parabola that satisfies conditions focus $\left(-5,3\right)$ and vertex $\left(-5,6\right)$ is ${\left(x+5\right)}^2=-12\left(y-6\right)$ .

Given information :

The parabola with focus $\left(-5,3\right)$ and vertex $\left(-5,6\right)$ .

Formula used :

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

Calculation :

Consider the parabola with focus $\left(-2,-4\right)$ and vertex $\left(-4,-4\right)$ ,

It is given that $\left(h,k\right)=\left(-5,6\right)$ and focus is at $\left(-5,3\right)$ .

Equate $k+p$ to $3$ and substitute $6$ for $k$ .

  $\begin{array}{c}k+p=3\\ 6+p=3\\ p=-3\end{array}$

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

  $\begin{array}{c}{\left(x-\left(-5\right)\right)}^2=4\left(-3\right)\left(y-6\right)\\ {\left(x+5\right)}^2=-12\left(y-6\right)\end{array}$

Thus, the equation in the standard form for the parabola that satisfies conditions focus $\left(-5,3\right)$ and vertex $\left(-5,6\right)$ is ${\left(x+5\right)}^2=-12\left(y-6\right)$ .