Chapter 9.1, Problem 78E

To determine

To find:The standard form of the equation of the parabola when the vertex is at $\left(3,-3\right)$ and focus is $\left(3,-\frac{9}{4}\right)$ .

Answer to Problem 78E

The standard form of the equation of the parabolais ${\left(x-3\right)}^2=3\left(y+3\right)$ .

Explanation of Solution

Given information:

The vertex of the parabola is at $\left(3,-3\right)$ and the focus is $\left(3,-\frac{9}{4}\right)$ .

Calculation:

The vertex of the parabola is at $\left(3,-3\right)$ and the focus is $\left(3,-\frac{9}{4}\right)$ . Both has the same x -coordinate. So, the parabola has a vertical axis.

Write the expression for the standard form of the parabola when it had ahorizontal axis.

  ${\left(x-h\right)}^2=4p\left(y-k\right)$

Where,

• $p>0$ .
• The vertex of the parabola is $\left(h,k\right)$ .

Calculate the value of $p$ from the given value of the focus.

  $\begin{array}{c}-3+p=-\frac{9}{4}\\ p=-\frac{9}{4}+3\\ p=\frac{3}{4}\end{array}$

Substitute the values in the standard equation of the parabola.

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

Therefore, the standard form of the equation of the parabolais ${\left(x-3\right)}^2=3\left(y+3\right)$ .