What is the vertices 16x^2-96x+9y^2=0 ?

The vertices 16x^2-96x+9y^2=0 is (3,4),(3,-4)

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vertices 16x^2-96x+9y^2=0

vertices

16 x^{2} - 96 x + 9 y^{2} = 0

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

Solution steps

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

Calculate ellipse properties

16 x^{2} - 96 x + 9 y^{2} = 0 :

Ellipse with center

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

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

Refine

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

4 x^{2} + 25 y^{2} - 50 y = 75

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

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

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

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

What is the vertices 16x^2-96x+9y^2=0 ?
The vertices 16x^2-96x+9y^2=0 is (3,4),(3,-4)