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

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

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

vertices

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

(8, - 1), (- 2, - 1)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(8, - 1), (- 2, - 1)

\frac{x ^{2}}{100} + \frac{y ^{2}}{36} = 1

x^{2} + 3 y^{2} = 84

16 x^{2} + 25 y^{2} + 32 x - 384 = 0

\frac{( x - 2 ) ^{2}}{64} + \frac{( y + 4 ) ^{2}}{16} = 1

25 x^{2} + 16 y^{2} + 100 x - 96 y - 156 = 0

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