What is the vertices 16x^2-96x+4y^2=-80 ?

The vertices 16x^2-96x+4y^2=-80 is (3,4),(3,-4)

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vertices 16x^2-96x+4y^2=-80

vertices

16 x^{2} - 96 x + 4 y^{2} = - 80

(3, 4), (3, - 4)

Solution steps

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

Calculate ellipse properties

16 x^{2} - 96 x + 4 y^{2} = - 80 :

Ellipse with center

(h, k) = (3, 0), b = 4, a = 2

(3, 0 + 4), (3, 0 - 4)

Refine

(3, 4), (3, - 4)

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

\frac{( x - 7 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{25} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

9 x^{2} + 4 y^{2} - 36 x + 40 y + 100 = 0

What is the vertices 16x^2-96x+4y^2=-80 ?
The vertices 16x^2-96x+4y^2=-80 is (3,4),(3,-4)