What is the vertices 16x^2+9y^2-64x-54y+1=0 ?

The vertices 16x^2+9y^2-64x-54y+1=0 is (2,7),(2,-1)

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vertices 16x^2+9y^2-64x-54y+1=0

vertices

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

(2, 7), (2, - 1)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0 :

Ellipse with center

(h, k) = (2, 3), b = 4, a = 3

(2, 3 + 4), (2, 3 - 4)

Refine

(2, 7), (2, - 1)

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

\frac{x ^{2}}{94} + \frac{y ^{2}}{36} = 1

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

What is the vertices 16x^2+9y^2-64x-54y+1=0 ?
The vertices 16x^2+9y^2-64x-54y+1=0 is (2,7),(2,-1)