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Q38.

Expert-verified
Found in: Page 424

### Algebra 2

Book edition Middle English Edition
Author(s) Carter
Pages 804 pages
ISBN 9780079039903

# Write an equation for the parabola having focus $\left(-4,-2\right)$ and directrix $x=-8$. Then draw the graph.

The required equation of the parabola is $x=\frac{1}{8}{\left(y+2\right)}^{2}-6$.

See the step by step solution

## Step 1. Write down the given information.

The given parabola has focus $\left(-4,-2\right)$ and directrix $x=-8$.

## Step 2. Concept used.

If $P\left(x,y\right)$ be any point on parabola having focus $\left({f}_{1},{f}_{2}\right)$ and directrix $x=a$ then:

Distance of point $P\left(x,y\right)$ from focus $\left({f}_{1},{f}_{2}\right)=$ Distance of point $P\left(x,y\right)$ from $\left(a,y\right)$

## Step 3. Calculation.

Since the given parabola has focus $\left(-4,-2\right)$ and directrix $x=-8$. Therefore, apply the concept stated above,

Distance of point $P\left(x,y\right)$ from focus $\left(-4,-2\right)$ Distance of point $P\left(x,y\right)$ from $\left(-8,y\right)$

$\begin{array}{c}\sqrt{{\left(x+4\right)}^{2}+{\left(y+2\right)}^{2}}=\sqrt{{\left(x+8\right)}^{2}+{\left(y-y\right)}^{2}}\\ {\left(x+4\right)}^{2}+{\left(y+2\right)}^{2}={\left(x+8\right)}^{2}+{\left(y-y\right)}^{2}....\left(\text{Squaring}\right)\\ {\left(x+4\right)}^{2}+{\left(y+2\right)}^{2}={\left(x+8\right)}^{2}\\ {\left(y+2\right)}^{2}={\left(x+8\right)}^{2}-{\left(x+4\right)}^{2}\\ {\left(y+2\right)}^{2}=\left(2x+12\right)\left(4\right)\\ {\left(y+2\right)}^{2}=8x+48\\ x=\frac{1}{8}{\left(y+2\right)}^{2}-6\end{array}$

Hence, $x=\frac{1}{8}{\left(y+2\right)}^{2}-6$ is the required equation of the parabola.

## Step 4. Sketch the graph of the parabola.

The graph of the parabola $x=\frac{1}{8}{\left(y+2\right)}^{2}-6$ is shown below.

## Step 5. Conclusion.

The required equation of the parabola is $x=\frac{1}{8}{\left(y+2\right)}^{2}-6$.