What is the vertices 4(x-5)^2+3(y-5)^2=192 ?

The vertices 4(x-5)^2+3(y-5)^2=192 is (5,13),(5,-3)

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vertices 4(x-5)^2+3(y-5)^2=192

vertices

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

(5, 13), (5, - 3)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192 :

Ellipse with center

(h, k) = (5, 5), b = 8, a = 43

(5, 5 + 8), (5, 5 - 8)

Refine

(5, 13), (5, - 3)

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

9 x^{2} + 36 x + y^{2} + 2 y + 36 = 0

27 x^{2} + y^{2} - 18 y = 0

\frac{( x - 2 ) ^{2}}{5} + \frac{( y + 3 ) ^{2}}{9} = 1

What is the vertices 4(x-5)^2+3(y-5)^2=192 ?
The vertices 4(x-5)^2+3(y-5)^2=192 is (5,13),(5,-3)