# Find the equation of parabola whose focus at (-3,0)the directrix X+5= 0 ?

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Jim G. Share
Aug 21, 2017

${y}^{2} - 4 x - 16 = 0$

#### Explanation:

$\text{from any point "(x,y)" on the parabola}$

$\text{the distance to the focus and the directrix from this point}$
$\text{are equal}$

$\textcolor{b l u e}{\text{using the distance formula then}}$

$\sqrt{{\left(x + 3\right)}^{2} + {\left(y - 0\right)}^{2}} = | x + 5 |$

$\textcolor{b l u e}{\text{squaring both sides}}$

${\left(x + 3\right)}^{2} + {y}^{2} = {\left(x + 5\right)}^{2}$

$\Rightarrow {x}^{2} + 6 x + 9 + {y}^{2} = {x}^{2} + 10 x + 25$

$\Rightarrow {y}^{2} - 4 x - 16 = 0 \text{ is the equation}$

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

8
Aug 21, 2017

${y}^{2} = 4 x + 16$

#### Explanation:

Given-
focus $\left(- 3 , 0\right)$
Directrix $x + 5 = 0$

Look at the diagram

General form of the equation

${y}^{2} = 4 a x$

Since vertex is away from the origin

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

Where (h,k) is vertex
From the given information

$h = - 4$
$k = 0$
$a = 1$

Substitute these values

${\left(y - 0\right)}^{2} = 4 \times 1 \times \left(x - \left(- 4\right)\right)$
${y}^{2} = 4 \left(x + 4\right)$
${y}^{2} = 4 x + 16$

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