What is the vertices 16(x-4)^2+25(y+2)^2=400 ?

The vertices 16(x-4)^2+25(y+2)^2=400 is (9,-2),(-1,-2)

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

vertices

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

(9, - 2), (- 1, - 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, - 2), a = 5, b = 4

(4 + 5, - 2), (4 - 5, - 2)

Refine

(9, - 2), (- 1, - 2)

3 x^{2} + 4 y^{2} - 12 x = 36

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

\frac{x ^{2}}{9} + \frac{( y + 5 ) ^{2}}{25} = 1

\frac{( x - 3 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{4} = 1

What is the vertices 16(x-4)^2+25(y+2)^2=400 ?
The vertices 16(x-4)^2+25(y+2)^2=400 is (9,-2),(-1,-2)