What is the vertices 9x^2-72x+y^2+4y+112=0 ?

The vertices 9x^2-72x+y^2+4y+112=0 is (4,4),(4,-8)

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vertices 9x^2-72x+y^2+4y+112=0

vertices

9 x^{2} - 72 x + y^{2} + 4 y + 112 = 0

(4, 4), (4, - 8)

Solution steps

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

Calculate ellipse properties

9 x^{2} - 72 x + y^{2} + 4 y + 112 = 0 :

Ellipse with center

(h, k) = (4, - 2), b = 6, a = 2

(4, - 2 + 6), (4, - 2 - 6)

Refine

(4, 4), (4, - 8)

4 x^{2} + 9 y^{2} = 1

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

36 x^{2} + 9 y^{2} + 48 x - 36 y + 43 = 0

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

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

What is the vertices 9x^2-72x+y^2+4y+112=0 ?
The vertices 9x^2-72x+y^2+4y+112=0 is (4,4),(4,-8)