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Q38.

Expert-verifiedFound in: Page 424

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** $x=\frac{1}{8}{\left(y+2\right)}^{2}-6$.**

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

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 $\left(a,y\right)$

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 $\left(-8,y\right)$

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

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

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

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

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