How do you find all the critical points to graph $x^{2} - 9 y^{2} + 2 x - 54 y + 80 = 0$ $x^2 - 9y^2 + 2x - 54y + 80 = 0$ including vertices, foci and asympotes?

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How do you find all the critical points to graph $x^{2} - 9 y^{2} + 2 x - 54 y + 80 = 0$ $x^2 - 9y^2 + 2x - 54y + 80 = 0$ including vertices, foci and asympotes?

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Answer 1 · 2016-10-19T18:20:18

Please see the explanation.

Explanation:

Add $h^{2} - 9 k^{2} - 80$ $h^2 - 9k^2 - 80$ to both sides:

$x^{2} + 2 x + h^{2} - 9 y^{2} - 54 y - 9 k^{2} = h^{2} - 9 k^{2} - 80$ $x^2 + 2x + h^2 - 9y^2 - 54y - 9k^2 = h^2 - 9k^2 - 80$

Set the right side of the pattern ${(x - h)}^{2} = x^{2} - 2 h x + h^{2}$ $(x - h)^2 = x^2 - 2hx + h^2$ equal to the terms $x^{2} + 2 x + h^{2}$ $x^2 + 2x + h^2$ :

$x^{2} - 2 h x + h^{2} = x^{2} + 2 x + h^{2}$ $x^2 - 2hx + h^2 = x^2 + 2x + h^2$

Solve for $h$ $h$ and $h^{2}$ $h^2$ :

$- 2 h x + h^{2} = 2 x + h^{2}$ $-2hx + h^2 = 2x + h^2$

$- 2 h x = 2 x$ $-2hx = 2x$

$h = - 1$ $h = -1$ and $h^{2} = 1$ $h^2 = 1$

Substitute the left side of the pattern (along with the value of h) for the 3 terms on the left and substitute 1 for $h^{2}$ $h^2$ on the right:

${(x - - 1)}^{2} - 9 y^{2} - 54 y - 9 k^{2} = 1 - 9 k^{2} - 80$ $(x - -1)^2 - 9y^2 - 54y - 9k^2 = 1 - 9k^2 - 80$

Factor -9 from the y terms:

${(x - 1)}^{2} - 9 (y^{2} + 6 y + k^{2}) = 1 - 9 k^{2} - 80$ $(x - 1)^2 - 9(y^2 + 6y + k^2) = 1 - 9k^2 - 80$

Set the right side of the pattern ${(y - k)}^{2} = y^{2} - 2 k y + k^{2}$ $(y - k)^2 = y^2 - 2ky + k^2$ equal to the terms $y^{2} + 6 y + k^{2}$ $y^2 + 6y + k^2$ :

$y^{2} - 2 k y + k^{2} = y^{2} + 6 y + k^{2}$ $y^2 - 2ky + k^2= y^2 + 6y + k^2$

Solve for $k$ $k$ and $k^{2}$ $k^2$ :

$- 2 k y = 6 y$ $-2ky= 6y$

$k = - 3$ $k = -3$ and $k^{2} = 9$ $k^2 = 9$

Substitute the left side on the pattern (along with the value of k) into the ()s on the left and 9 for $k^{2}$ $k^2$ on the right:

${(x - - 1)}^{2} - 9 ({(y - - 3)}^{2}) = 1 - 9 (9) - 80$ $(x - -1)^2 - 9((y - -3)^2) = 1 - 9(9) - 80$

Simplify the right side:

${(x - - 1)}^{2} - 9 ({(y - - 3)}^{2}) = - 160$ $(x - -1)^2 - 9((y - -3)^2) = -160$

Divide both sides by -160:

$9 \frac{{(y - - 3)}^{2}}{160} - \frac{{(x - - 1)}^{2}}{160} = 1$ $9((y - -3)^2)/160 - (x - -1)^2/160 = 1$

Write denominators as squares:

$\frac{{(y - - 3)}^{2}}{{(\frac{4 \sqrt{10}}{3})}^{2}} - \frac{{(x - - 1)}^{2}}{{(4 \sqrt{10})}^{2}} = 1$ $(y - -3)^2/((4sqrt(10))/3)^2 - (x - -1)^2/(4sqrt10)^2 = 1$

Center: $(- 1, - 3)$ $(-1,-3)$
To find the vertices, make the negative term disappear by setting $x = - 1$ $x = -1$ :

$\frac{{(y - - 3)}^{2}}{{(\frac{4 \sqrt{10}}{3})}^{2}} = 1$ $(y - -3)^2/((4sqrt(10))/3)^2 = 1$

${(y - - 3)}^{2} = {(\frac{4 \sqrt{10}}{3})}^{2}$ $(y - -3)^2 = ((4sqrt(10))/3)^2$

$y - - 3 = \frac{4 \sqrt{10}}{3}$ $y - -3 = (4sqrt(10))/3$ and $y - - 3 = - \frac{4 \sqrt{10}}{3}$ $y - -3 = -(4sqrt(10))/3$

$y = - 3 + \frac{4 \sqrt{10}}{3}$ $y = -3 + (4sqrt(10))/3$ and $y = - 3 - \frac{4 \sqrt{10}}{3}$ $y = -3 -(4sqrt(10))/3$
Vertices: $(- 1, - 3 + \frac{4 \sqrt{10}}{3})$ $(-1, -3 + (4sqrt(10))/3)$ and $(- 1, - 3 - \frac{4 \sqrt{10}}{3})$ $(-1, -3 -(4sqrt(10))/3)$

To find the focal distance, c, take the square root of the sum of the squares of the denominators:

$c = \sqrt{\frac{160}{9} + 160}$ $c = sqrt(160/9 + 160)$

$c = \frac{40}{3}$ $c = 40/3$

The foci are at : $(- 1, - 3 + \frac{40}{3})$ $(-1, -3 + 40/3)$ and $(- 1, - 3 - \frac{40}{3})$ $(-1, -3 - 40/3)$

Foci: $(- 1, \frac{31}{3})$ $(-1, 31/3)$ and $(- 1, - \frac{49}{3})$ $(-1, - 49/3)$

To find the asymptotes, please observe that dividing the first denominator by the second gives the slopes of $\frac{1}{3}$ $1/3$ and $- \frac{1}{3}$ $-1/3$ . All that is left to do is to force the lines with those two slopes, through the center point.

$y = \frac{1}{3} (x - - 1) - 3$ $y = 1/3(x - -1) - 3$
$y = - \frac{1}{3} (x - - 1) - 3$ $y = -1/3(x - -1) - 3$

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How do you find all the critical points to graph $x^{2} - 9 y^{2} + 2 x - 54 y + 80 = 0$ $x^2 - 9y^2 + 2x - 54y + 80 = 0$ including vertices, foci and asympotes?

1 Answer

Explanation:

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How do you find all the critical points to graph x2−9y2+2x−54y+80=0x^2 - 9y^2 + 2x - 54y + 80 = 0 including vertices, foci and asympotes?

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How do you find all the critical points to graph $x^{2} - 9 y^{2} + 2 x - 54 y + 80 = 0$ $x^2 - 9y^2 + 2x - 54y + 80 = 0$ including vertices, foci and asympotes?