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

The vertices 16(x-1)^2+25(y-4)^2=400 is (6,4),(-4,4)

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vertices 16(x-1)^2+25(y-4)^2=400

vertices

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

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

Solution steps

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

Calculate ellipse properties

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400 :

Ellipse with center

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

(1 + 5, 4), (1 - 5, 4)

Refine

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

25 x^{2} + 16 y^{2} - 50 x + 64 y - 311 = 0

16 x^{2} + 36 y^{2} - 256 x + 448 = 0

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

3 x^{2} + 4 y^{2} = 48

What is the vertices 16(x-1)^2+25(y-4)^2=400 ?
The vertices 16(x-1)^2+25(y-4)^2=400 is (6,4),(-4,4)