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

The vertices ((x-3)^2)/(16)+((y+1)^2)/(25)=1 is (3,4),(3,-6)

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

vertices

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

(3, 4), (3, - 6)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (3, - 1), b = 5, a = 4

(3, - 1 + 5), (3, - 1 - 5)

Refine

(3, 4), (3, - 6)

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

\frac{( x + 6 ) ^{2}}{36} + \frac{( y - 7 ) ^{2}}{32} = 1

16 x^{2} + 4 y^{2} + 96 x - 24 y + 116 = 0

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

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

What is the vertices ((x-3)^2)/(16)+((y+1)^2)/(25)=1 ?
The vertices ((x-3)^2)/(16)+((y+1)^2)/(25)=1 is (3,4),(3,-6)