How do you find the vertex, focus and directrix of $x^{2} - 2 x + 44 y + 353 = 0$ $x^2 - 2x + 44y + 353 = 0$ ?

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Answer 1 · 2017-05-16T21:39:15

Vertex: $(1, - 8)$ $(1, -8)$
Focus: $(1, - 19)$ $(1, -19)$
Directrix: $y = 3$ $y = 3$

Start by moving the constant and $y$ $y$ -value to the right side of the equation, then complete the square for $x$ $x$ :

$(x^{2} - 2 x + 1) = - 44 y - 353 + 1$ $(x^2 - 2x + 1) = -44y - 353 + 1$

${(x - 1)}^{2} = - 44 y - 352$ $(x-1)^2 = -44y - 352$

$- \frac{1}{44} {(x - 1)}^{2} = y + 8$ $-1/44(x-1)^2 = y + 8$

We can now see that the vertex is $(1, - 8)$ $(1, -8)$

The focus is $(h, k + \frac{1}{4 a})$ $(h, k + 1/(4a))$ , which makes the focus of this equation

$(1, - 8 + \frac{1}{4 (- \frac{1}{44})}) = (1, - 19)$ $(1, -8 + 1/(4(-1/44))) = (1, -19)$

The directrix of this equation is

$y = k - \frac{1}{4 a} = - 8 - \frac{1}{4 (- \frac{1}{44})} = 3$ $y = k - 1/(4a) = -8 - 1/(4(-1/44)) = 3$

How do you find the vertex, focus and directrix of x^2 - 2x + 44y + 353 = 0 x2−2x+44y+353=0x^2 - 2x + 44y + 353 = 0 ?