Chapter 10, Problem 1MC

To determine

Whether the statement “The graph of ${\left(x+3\right)}^2=8\left(y-2\right)$ is a parabola with vertex $\left(-3,2\right)$” is true or false.

Answer to Problem 1MC

The statement is true.

Explanation of Solution

Definitions used:

1) The standard equation of a parabola with vertex $\left(h,k\right)$ and the vertical axis of symmetry is ${\left(x-h\right)}^2=4p\left(y-k\right)$, where the vertex is $\left(h,k\right)$ the focus is $\left(h,k+p\right)$ and the directrix is $y=k-p$.

2) The standard equation of a parabola with vertex $\left(h,k\right)$ and the horizontal axis of symmetry is ${\left(y-k\right)}^2=4p\left(x-h\right)$, where the vertex is $\left(h,k\right)$ the focus is $\left(h+p,k\right)$ and the directrix is $x=h-p$.

Calculation:

The given equation is ${\left(x+3\right)}^2=8\left(y-2\right)$.

Convert the equation ${\left(x+3\right)}^2=8\left(y-2\right)$ into standard form as follows.

$\begin{array}{l}{\left(x+3\right)}^2=8\left(y-2\right)\\ {\left(x-\left(-3\right)\right)}^2=4\cdot 2\left(y-2\right)\end{array}$

Compare ${\left(x-\left(-3\right)\right)}^2=4\cdot 2\left(y-2\right)$ with general equation of parabola ${\left(x-h\right)}^2=4p\left(y-k\right)$.

Here, $h=-3,k=2$.

Thus, the vertex is $\left(-3,2\right)$.

So, the given equation ${\left(x+3\right)}^2=8\left(y-2\right)$ is in the form of parabola.

Therefore, the given statement is true.