How do you find the coordinates of the center, foci, the length of the major and minor axis given $3 x^{2} + y^{2} + 18 x - 2 y + 4 = 0$ $3x^2+y^2+18x-2y+4=0$ ?

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How do you find the coordinates of the center, foci, the length of the major and minor axis given $3 x^{2} + y^{2} + 18 x - 2 y + 4 = 0$ $3x^2+y^2+18x-2y+4=0$ ?

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Answer 1 · 2017-01-10T00:21:51

Given: $\frac{{(y - k)}^{2}}{a^{2}} + \frac{{(x - h)}^{2}}{b^{2}} = 1 [1]$ $(y - k)^2/a^2 + (x - h)^2/b^2 = 1" [1]"$
Center: $(h, k)$ $(h,k)$
Foci: $(h, k - \sqrt{a^{2} - b^{2}}) and (h, k + \sqrt{a^{2} - b^{2}})$ $(h, k-sqrt(a^2 - b^2)) and (h, k+sqrt(a^2 - b^2))$
major axis = 2a
minor axis = 2b

Explanation:

The following are the steps to put the given equation into the form of equation [1]:

Subtract 4 from both sides:

$3 x^{2} + y^{2} + 18 x - 2 y = - 4 [2]$ $3x^2 + y^2 + 18x - 2y = - 4" [2]"$

Group the x terms and the y terms together on the left:

$3 x^{2} + 18 x + y^{2} - 2 y = - 4 [3]$ $3x^2 + 18x + y^2 - 2y = - 4" [3]"$

Because the coefficient of the x^2 term is 3, add #3h^2 to both sides ; make it the 3rd term on the left and the first term on the right:

$3 x^{2} + 18 x + 3 h^{2} + y^{2} - 2 y = 3 h^{2} - 4 [4]$ $3x^2 + 18x + 3h^2 + y^2 - 2y = 3h^2 - 4" [4]"$

Because the coefficient of the y^2 term is 1, add k^2 to both sides; make it the sixth term on the left and the second term on the right:

$3 x^{2} + 18 x + 3 h^{2} + y^{2} - 2 y + k^{2} = 3 h^{2} + k^{2} - 4 [5]$ $3x^2 + 18x + 3h^2 + y^2 - 2y + k^2 = 3h^2 + k^2 - 4" [5]"$

Remove a common factor of 3 from the first 3 terms:

$3 (x^{2} + 6 x + h^{2}) + y^{2} - 2 y + k^{2} = 3 h^{2} + k^{2} - 4 [6]$ $3(x^2 + 6x + h^2) + y^2 - 2y + k^2 = 3h^2 + k^2 - 4" [6]"$

Use the pattern for ${(x - h)}^{2} = x^{2} - 2 h x + h^{2}$ $(x - h)^2 = x^2 - 2hx + h^2$ .

Match the "-2hx" term in the pattern with the "6x" term in equation [6] and write the equation:

$- 2 h x = 6 x$ $-2hx = 6x$

Solve for h:

$h = - 3$ $h = -3$

This means that we can substitute ${(x - - 3)}^{2}$ $(x - -3)^2$ for $(x^{2} + 6 x + h^{2})$ $(x^2 + 6x + h^2)$ on the left side of equation [6] and -3 for h on the right:

$3 {(x - - 3)}^{2} + y^{2} - 2 y + k^{2} = 3 {(- 3)}^{2} + k^{2} - 4 [7]$ $3(x - -3)^2 + y^2 - 2y + k^2 = 3(-3)^2 + k^2 - 4" [7]"$

Use the pattern for ${(y - k)}^{2} = y^{2} - 2 k y + k^{2}$ $(y - k)^2 = y^2 - 2ky + k^2$ .

Match the "-2ky" term in the pattern with the "-2y" term in equation [7] and write the equation:

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

Solve for k:

$k = 1$ $k = 1$

This means that we can substitute ${(y - 1)}^{2}$ $(y - 1)^2$ for $y^{2} - 2 y + k^{2}$ $y^2 -2y + k^2$ on the left side of equation [7] and 1 for k on the right:

$3 {(x - - 3)}^{2} + {(y - 1)}^{2} = 3 {(- 3)}^{2} + 1^{2} - 4 [8]$ $3(x - -3)^2 + (y - 1)^2 = 3(-3)^2 + 1^2 - 4" [8]"$

Simplify the right:

$3 {(x - - 3)}^{2} + {(y - 1)}^{2} = 6 [9]$ $3(x - -3)^2 + (y - 1)^2 = 6" [9]"$

Divide both sides by 6:

$\frac{{(x - - 3)}^{2}}{2} + \frac{{(y - 1)}^{2}}{6} = 1 [10]$ $(x - -3)^2/2 + (y - 1)^2/6 = 1" [10]"$

Swap terms and write the denominators as squares:

$\frac{{(y - 1)}^{2}}{{(\sqrt{6})}^{2}} + \frac{{(x - - 3)}^{2}}{{(\sqrt{2})}^{2}} = 1 [11]$ $(y - 1)^2/(sqrt6)^2+(x - -3)^2/(sqrt2)^2 = 1" [11]"$

We have the form of equation [1]

$h = - 3$ $h = -3$
$k = 1$ $k = 1$
$a = \sqrt{6}$ $a = sqrt6$
$b = \sqrt{2}$ $b = sqrt2$
$\sqrt{a^{2} - b^{2}} = \sqrt{6 - 2} = \sqrt{4} = 2$ $sqrt(a^2 - b^2) = sqrt(6 - 2) = sqrt(4) = 2$

Center: $(- 3, 1)$ $(-3, 1)$
Foci: $(- 3, 1 - 2) and (- 3, 1 + 2) = (- 3, - 1) and (- 3, 3)$ $(-3, 1-2) and (-3, 1 + 2) = (-3, -1) and (-3, 3)$
Major axis: $2 \sqrt{6}$ $2sqrt6$
Minor axis: $2 \sqrt{2}$ $2sqrt2$

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How do you find the coordinates of the center, foci, the length of the major and minor axis given $3 x^{2} + y^{2} + 18 x - 2 y + 4 = 0$ $3x^2+y^2+18x-2y+4=0$ ?

1 Answer

Explanation:

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How do you find the coordinates of the center, foci, the length of the major and minor axis given 3x2+y2+18x−2y+4=03x^2+y^2+18x-2y+4=0?

1 Answer

Explanation:

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How do you find the coordinates of the center, foci, the length of the major and minor axis given $3 x^{2} + y^{2} + 18 x - 2 y + 4 = 0$ $3x^2+y^2+18x-2y+4=0$ ?