How do you find the vertex, focus, and directrix of the parabola $y^{2} + 6 y + 8 x + 25 = 0$ $y^2+6y+8x+25=0$ ?

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Answer 1 · 2017-02-06T07:35:09

The vertex is $= (- 2, - 3)$ $=(-2,-3)$
The focus is $= (- 4, - 3)$ $=(-4,-3)$
The directrix is $x = 0$ $x=0$

Rewrite the equation and complete the squares

$y^{2} + 6 y + 8 x + 25 = 0$ $y^2+6y+8x+25=0$

$y^{2} + 6 y = - 8 x - 25$ $y^2+6y=-8x-25$

$y^{2} + 6 y + 9 = - 8 x - 25 + 9$ $y^2+6y+9=-8x-25+9$

${(y + 3)}^{2} = - 8 x - 16$ $(y+3)^2=-8x-16$

${(y + 3)}^{2} = - 8 (x + 2)$ $(y+3)^2=-8(x+2)$

We compare this equation to

${(y - b)}^{2} = 2 p (x - a)$ $(y-b)^2=2p(x-a)$

$2 p = - 8$ $2p=-8$ , $\Rightarrow$ $=>$ , $p = - 4$ $p=-4$

The vertex is $(a, b) = (- 2, - 3)$ $(a,b)=(-2,-3)$

The focus is $(a + \frac{p}{2}, b) = (- 4, - 3)$ $(a+p/2,b)=(-4,-3)$

The directrix is $x = a - \frac{p}{2}$ $x=a-p/2$ , $\Rightarrow$ $=>$ , $x = - 2 + 2 = 0$ $x=-2+2=0$

graph{(y^2+6y+8x+25)((x+2)^2+(y+3)^2-0.1)((x+4)^2+(y+3)^2-0.1)=0 [-59.95, 13.1, -21.5, 15.05]}

How do you find the vertex, focus, and directrix of the parabola y^2+6y+8x+25=0y2+6y+8x+25=0y^2+6y+8x+25=0?