What is the vertices 16x^2-32x+9y^2=128 ?

The vertices 16x^2-32x+9y^2=128 is (1,4),(1,-4)

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vertices 16x^2-32x+9y^2=128

vertices

16 x^{2} - 32 x + 9 y^{2} = 128

(1, 4), (1, - 4)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

16 x^{2} - 32 x + 9 y^{2} = 128 :

Ellipse with center

(h, k) = (1, 0), b = 4, a = 3

(1, 0 + 4), (1, 0 - 4)

Refine

(1, 4), (1, - 4)

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

5 x^{2} + 2 y^{2} = 10

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

4 x^{2} + 16 y^{2} - 64 = 0

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

What is the vertices 16x^2-32x+9y^2=128 ?
The vertices 16x^2-32x+9y^2=128 is (1,4),(1,-4)