What is the vertices ((x-4)^2)/(25)+((y+2)^2)/(64)=1 ?

The vertices ((x-4)^2)/(25)+((y+2)^2)/(64)=1 is (4,6),(4,-10)

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vertices ((x-4)^2)/(25)+((y+2)^2)/(64)=1

vertices

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{64} = 1

(4, 6), (4, - 10)

Solution steps

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

Calculate ellipse properties

\frac{( x - 4 ) ^{2}}{25} + \frac{( y + 2 ) ^{2}}{64} = 1 :

Ellipse with center

(h, k) = (4, - 2), b = 8, a = 5

(4, - 2 + 8), (4, - 2 - 8)

Refine

(4, 6), (4, - 10)

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

\frac{( x - 2 ) ^{2}}{16} + (y - 5)^{2} = 1

\frac{x ^{2}}{16} + \frac{( y - 4 ) ^{2}}{1} = 1

\frac{x ^{2}}{56} + \frac{y ^{2}}{36} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} - 400 = 0

What is the vertices ((x-4)^2)/(25)+((y+2)^2)/(64)=1 ?
The vertices ((x-4)^2)/(25)+((y+2)^2)/(64)=1 is (4,6),(4,-10)