# What are the vertex, focus and directrix of  9y=x^2-2x+9 ?

Nov 21, 2017

Vertex $\left(1 , \frac{8}{9}\right)$
Focus $\left(1 , \frac{113}{36}\right)$
Directrix $y = - \frac{49}{36}$

#### Explanation:

Given -

$9 y = {x}^{2} - 2 x + 9$

vertex?
Focus ?
Directrix?

${x}^{2} - 2 x + 9 = 9 y$

To find Vertex, Focus and directrix, we have to rewrite the given equation in vertex form i.e., ${\left(x - h\right)}^{2} = 4 a \left(y - k\right)$

${x}^{2} - 2 x = 9 y - 9$

${x}^{2} - 2 x + 1 = 9 y - 9 + 1$

${\left(x - 1\right)}^{2} = 9 y - 8$

${\left(x - 1\right)}^{2} = 9 \left(y - \frac{8}{9}\right)$

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To find the equation in terms of $y$ [This not asked in the problem]

$9 \left(y - \frac{8}{9}\right) = {\left(x - 1\right)}^{2}$

$y - \frac{8}{9} = \frac{1}{9.} {\left(x - 1\right)}^{2}$

$y = \frac{1}{9.} {\left(x - 1\right)}^{2} + \frac{8}{9}$

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Let us uses $9 \left(y - \frac{8}{9}\right) = {\left(x - 1\right)}^{2}$ to find the vertex, focus and directrix.

${\left(x - 1\right)}^{2} = 4 \times \frac{9}{4} \left(y - \frac{8}{9}\right)$

Vertex $\left(1 , \frac{8}{9}\right)$

Focus $\left(1 , \left(\frac{8}{9} + \frac{9}{4}\right)\right)$
Focus $\left(1 , \frac{113}{36}\right)$
Directrix $y = \frac{8}{9} - \frac{9}{4}$
Directrix $y = - \frac{49}{36}$