What is the vertices ((x+1)^2)/(64)+((y-6)^2)/(16)=1 ?

The vertices ((x+1)^2)/(64)+((y-6)^2)/(16)=1 is (7,6),(-9,6)

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vertices ((x+1)^2)/(64)+((y-6)^2)/(16)=1

vertices

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

(7, 6), (- 9, 6)

Solution steps

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

Calculate ellipse properties

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 1, 6), a = 8, b = 4

(- 1 + 8, 6), (- 1 - 8, 6)

Refine

(7, 6), (- 9, 6)

\frac{x ^{2}}{100} + \frac{y ^{2}}{25} = 1

12.92 x^{2} + 0.07 y^{2} - 36 = 0

15 x^{2} + 16 y^{2} = 240

\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{16} = 1

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

What is the vertices ((x+1)^2)/(64)+((y-6)^2)/(16)=1 ?
The vertices ((x+1)^2)/(64)+((y-6)^2)/(16)=1 is (7,6),(-9,6)