What is the vertices ((x-6)^2)/(16)+((y+3)^2)/(25)=1 ?

The vertices ((x-6)^2)/(16)+((y+3)^2)/(25)=1 is (6,2),(6,-8)

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vertices ((x-6)^2)/(16)+((y+3)^2)/(25)=1

vertices

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

(6, 2), (6, - 8)

Solution steps

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

Calculate ellipse properties

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (6, - 3), b = 5, a = 4

(6, - 3 + 5), (6, - 3 - 5)

Refine

(6, 2), (6, - 8)

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

16 x^{2} - 32 x + 9 y^{2} = 128

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

5 x^{2} + 2 y^{2} = 10

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

What is the vertices ((x-6)^2)/(16)+((y+3)^2)/(25)=1 ?
The vertices ((x-6)^2)/(16)+((y+3)^2)/(25)=1 is (6,2),(6,-8)