Chapter A5.1, Problem 9E

To determine

To find: the given parabolas focus and directrix.

Answer to Problem 9E

The given parabolas focus is $(0,p)=(0,\frac{-3}{2})$ and directrix $\text{y =}\frac{3}{2}$ .

Explanation of Solution

Given:

  $x^2=-6y$

Calculation:

The given parabolas focus and directrix is calculated below

  $x^2=-6y$

We find the value of p in the standard equation

  $\begin{array}{l}{x}^2=4py\\ 4p=-6\\ p=\frac{-3}{2}\end{array}$

focus : $(0,p)=(0,\frac{-3}{2})$

directrix : $y=-p\text{ or y =}\frac{3}{2}$

Conclusion:

Therefore, the given parabolas focus is $(0,p)=(0,\frac{-3}{2})$ and directrix $\text{y =}\frac{3}{2}$ .