How do you put $6 x^{2} + 5 y^{2} - 24 x + 20 y + 14 = 0$ $6x^2+5y^2-24x+20y+14=0$ in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

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How do you put $6 x^{2} + 5 y^{2} - 24 x + 20 y + 14 = 0$ $6x^2+5y^2-24x+20y+14=0$ in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

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Answer 1 · 2017-03-15T22:11:05

Standard forms for the equation of an ellipse are $\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2+(y-k)^2/b^2=1$ or $\frac{{(y - k)}^{2}}{a^{2}} + \frac{{(x - h)}^{2}}{b^{2}} = 1; a > b$ $(y-k)^2/a^2+(x-h)^2/b^2=1; a > b$

Explanation:

Reference for an ellipse
Given: $6 x^{2} + 5 y^{2} - 24 x + 20 y + 14 = 0$ $6x^2+5y^2-24x+20y+14=0$

Add $6 h^{2} + 5 k^{2} - 14$ $6h^2 + 5k^2 - 14$ to both sides of the equation:

$6 x^{2} - 24 x + 6 h^{2} + 5 y^{2} + 20 y + 5 k^{2} = 6 h^{2} + 5 k^{2} - 14$ $6x^2-24x+6h^2+5y^2+20y+5k^2=6h^2+5k^2-14$

Remove a common factor of 6 from the first 3 terms and a common factor of 5 from the next 3 terms:

$6 (x^{2} - 4 x + h^{2}) + 5 (y^{2} + 4 y + k^{2}) = 6 h^{2} + 5 k^{2} - 14 [1]$ $6(x^2-4x+h^2)+5(y^2+4y+k^2)=6h^2+5k^2-14" [1]"$

To find the value of h, set the middle term of 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 middle term in the first parenthesis in equation [1]:

$- 2 h x = - 4 x$ $-2hx = -4x$

$h = 2$ $h = 2$

Substitute the left side of the pattern into equation [1]:

$6 {(x - h)}^{2} + 5 (y^{2} + 4 y + k^{2}) = 6 h^{2} + 5 k^{2} - 14 [2]$ $6(x - h)^2+5(y^2+4y+k^2)=6h^2+5k^2-14" [2]"$

Substitute 2 for h everywhere in equation [2]:

$6 {(x - 2)}^{2} + 5 (y^{2} + 4 y + k^{2}) = 6 {(2)}^{2} + 5 k^{2} - 14 [3]$ $6(x - 2)^2+5(y^2+4y+k^2)=6(2)^2+5k^2-14" [3]"$

To find the value of k, set the middle term of 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 middle term in the second parenthesis in equation [3]:

$- 2 k y = 4 y$ $-2ky = 4y$

$k = - 2$ $k = -2$

Substitute the left side of the pattern into equation [3]:

$6 {(x - 2)}^{2} + 5 {(y - k)}^{2} = 6 {(2)}^{2} + 5 k^{2} - 14 [4]$ $6(x - 2)^2+5(y - k)^2=6(2)^2+5k^2-14" [4]"$

Substitute -2 for k everywhere in equation [4]:

$6 {(x - 2)}^{2} + 5 {(y - - 2)}^{2} = 6 {(2)}^{2} + 5 {(- 2)}^{2} - 14 [5]$ $6(x - 2)^2+5(y - -2)^2=6(2)^2+5(-2)^2-14" [5]"$

Simplify the right side of equation [5]:

$6 {(x - 2)}^{2} + 5 {(y - - 2)}^{2} = 30 [6]$ $6(x - 2)^2+5(y - -2)^2=30" [6]"$

Divide both sides of the equation by 30:

$\frac{{(x - 2)}^{2}}{5} + \frac{{(y - - 2)}^{2}}{6} = 1 [7]$ $(x - 2)^2/5+(y - -2)^2/6=1" [7]"$

Make the denominators squares and swap terms:

$\frac{{(y - - 2)}^{2}}{{(\sqrt{6})}^{2}} + \frac{{(x - 2)}^{2}}{{(\sqrt{5})}^{2}} = 1 [8]$ $(y - -2)^2/(sqrt(6))^2+(x - 2)^2/(sqrt5)^2=1" [8]"$

Equation [8] is the standard form where, $h = 2, k = - 2, a = \sqrt{6}, and b = \sqrt{5}$ $h = 2, k = -2, a = sqrt(6), and b = sqrt(5)$

From the reference:

$c = \sqrt{a^{2} - b^{2}}$ $c = sqrt(a^2-b^2)$

$c = \sqrt{6 - 5}$ $c = sqrt(6 - 5)$

$c = 1$ $c = 1$

The center is at $(h, k)$ $(h,k)$ :

$(2, - 2)$ $(2,-2)$

The vertices are at #(h, k-a) and (h,k+a):

$(2, - 2 - \sqrt{6}) and (2, - 2 + \sqrt{6})$ $(2,-2-sqrt(6)) and (2,-2+sqrt(6))$

The foci are at: #(h, k-c) and (h,k+c):

$(2, - 2 - 1) and (2, - 2 + 1)$ $(2,-2-1) and (2, -2+1)$

$(2, - 3) and (2, - 1)$ $(2,-3) and (2,-1)$

The co-vertices are at $(h - b, k) and (h + b, k)$ $(h-b, k) and (h+b,k)$ :

$(2 - \sqrt{5}, - 2) and (2 + \sqrt{5}, - 2)$ $(2-sqrt(5),-2) and (2+sqrt(5),-2)$

The eccentricity is $\frac{c}{a} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$ $c/a = 1/sqrt(6)= sqrt(6)/6$

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How do you put $6 x^{2} + 5 y^{2} - 24 x + 20 y + 14 = 0$ $6x^2+5y^2-24x+20y+14=0$ in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

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Explanation:

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How do you put $6 x^{2} + 5 y^{2} - 24 x + 20 y + 14 = 0$ $6x^2+5y^2-24x+20y+14=0$ in standard form, find the center, the endpoints, vertices, the foci and eccentricity?