Chapter 9, Problem 22CR

To determine

To find: the standard form of equation of parabola.

Answer to Problem 22CR

The standard form of parabola is ${\left(x+6\right)}^2=-9\left(y-4\right)$ .

Explanation of Solution

Given information:

From the figure provided in the question it can be observed that,

Vertex: $\left(-6,4\right)$ , point on parabola is $\left(0,0\right)$ and parabola opens at the downwards.

Calculation: to find the standard form of parabola,

Since, $\left(-6,4\right)$ , point on parabola is $\left(0,0\right)$ and parabola opens at the downwards.

Thus, the standard form of parabola should be ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

Since, vertex is $\left(-6,4\right)$ , so $h=-6,k=4$

Thus, the equation is ,

  $\begin{array}{l}{\left(x-h\right)}^2=4p\left(y-k\right)\\ {\left(x+6\right)}^2=4p\left(y-4\right)\text{ [substitute and simplify]}\end{array}$

Now, point at parabola is $\left(0,0\right)$ , now substitute and simplify,

  $\begin{array}{l}{\left(x+6\right)}^2=4p\left(y-4\right)\\ {\left(0+6\right)}^2=4p\left(0-4\right)\text{ [substitute]}\\ 36=-16p\text{ [multiply]}\\ p=-\frac{9}{4}\text{ [divide]}\end{array}$

Thus, the equation is ${\left(x+6\right)}^2=-9\left(y-4\right)$ .