What is the vertices ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?

The vertices ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is (10,-7),(0,-7)

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vertices ((x-5)^2)/(25)+((y+7)^2)/(16)=1

vertices

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

(10, - 7), (0, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (5, - 7), a = 5, b = 4

(5 + 5, - 7), (5 - 5, - 7)

Refine

(10, - 7), (0, - 7)

(\frac{x ^{2}}{15}) + (\frac{y ^{2}}{16}) = 1

9 x^{2} - 36 x + 4 y^{2} = 0

x^{2} + 4 y^{2} = 4

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1

8 x^{2} + 2 y^{2} + 16 x = 0

What is the vertices ((x-5)^2)/(25)+((y+7)^2)/(16)=1 ?
The vertices ((x-5)^2)/(25)+((y+7)^2)/(16)=1 is (10,-7),(0,-7)