What is the vertices ((x-3)^2)/(64)+((y+5)^2)/(81)=1 ?

The vertices ((x-3)^2)/(64)+((y+5)^2)/(81)=1 is (3,4),(3,-14)

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vertices ((x-3)^2)/(64)+((y+5)^2)/(81)=1

vertices

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1

(3, 4), (3, - 14)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1 :

Ellipse with center

(h, k) = (3, - 5), b = 9, a = 8

(3, - 5 + 9), (3, - 5 - 9)

Refine

(3, 4), (3, - 14)

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

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

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

4 x^{2} + 25 y^{2} - 100 = 0

16 x^{2} + 25 y^{2} - 400 = 0

What is the vertices ((x-3)^2)/(64)+((y+5)^2)/(81)=1 ?
The vertices ((x-3)^2)/(64)+((y+5)^2)/(81)=1 is (3,4),(3,-14)