What is the vertices (2(x-6)^2)/3+(7(y+4)^2)/4 =1x ?

The vertices (2(x-6)^2)/3+(7(y+4)^2)/4 =1x is ((27+3sqrt(17))/4 ,-4),((27-3sqrt(17))/4 ,-4)

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vertices (2(x-6)^2)/3+(7(y+4)^2)/4 =1x

vertices

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1 x

(\frac{27 + 3 17}{4}, - 4), (\frac{27 - 3 17}{4}, - 4)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1 x :

Ellipse with center

(h, k) = (\frac{27}{4}, - 4), a = \frac{3 17}{4}, b = \frac{51}{14}

(\frac{27}{4} + \frac{3 17}{4}, - 4), (\frac{27}{4} - \frac{3 17}{4}, - 4)

Refine

(\frac{27 + 3 17}{4}, - 4), (\frac{27 - 3 17}{4}, - 4)

\frac{x ^{2}}{12} + \frac{y ^{2}}{144} = \frac{y}{12}

81 x^{2} + 36 y^{2} = 2916

(x)^{2} + \frac{( y - 1 ) ^{2}}{4} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

25 x^{2} + 16 y^{2} - 50 x + 64 y - 311 = 0

What is the vertices (2(x-6)^2)/3+(7(y+4)^2)/4 =1x ?
The vertices (2(x-6)^2)/3+(7(y+4)^2)/4 =1x is ((27+3sqrt(17))/4 ,-4),((27-3sqrt(17))/4 ,-4)