What is the vertices ((x-2)^2)/(64)+((y+4)^2)/(16)=1 ?

The vertices ((x-2)^2)/(64)+((y+4)^2)/(16)=1 is (10,-4),(-6,-4)

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vertices ((x-2)^2)/(64)+((y+4)^2)/(16)=1

vertices

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

(10, - 4), (- 6, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, - 4), a = 8, b = 4

(2 + 8, - 4), (2 - 8, - 4)

Refine

(10, - 4), (- 6, - 4)

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

x^{2} + 4 y^{2} - 6 x - 16 y + 21 = 0

100 x^{2} + 36 y^{2} = 225

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1

What is the vertices ((x-2)^2)/(64)+((y+4)^2)/(16)=1 ?
The vertices ((x-2)^2)/(64)+((y+4)^2)/(16)=1 is (10,-4),(-6,-4)