What is the vertices 4y^2-x^2+2x+16y+7=0 ?

The vertices 4y^2-x^2+2x+16y+7=0 is (1,-2+sqrt(2)),(1,-2-sqrt(2))

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vertices 4y^2-x^2+2x+16y+7=0

vertices

4 y^{2} - x^{2} + 2 x + 16 y + 7 = 0

(1, - 2 + 2), (1, - 2 - 2)

Solution steps

(h, k + a), (h, k - a)

Calculate Hyperbola properties

4 y^{2} - x^{2} + 2 x + 16 y + 7 = 0 :

Up-down Hyperbola with

(h, k) = (1, - 2), a = 2, b = 22

(1, - 2 + 2), (1, - 2 - 2)

4 x^{2} - 9 y^{2} + 32 x + 36 y + 64 = 0

- 16 x^{2} + 9 y^{2} = 144

x^{2} - y^{2} + 4 x - 6 y = 6

\frac{x ^{2}}{25} - \frac{y ^{2}}{4} = 1

x^{2} - 4 y^{2} + 2 x + 8 y - 7 = 0

What is the vertices 4y^2-x^2+2x+16y+7=0 ?
The vertices 4y^2-x^2+2x+16y+7=0 is (1,-2+sqrt(2)),(1,-2-sqrt(2))