What is the foci 0=-400/9 x^2+16y^2+400 ?

The foci 0=-400/9 x^2+16y^2+400 is (sqrt(34),0),(-sqrt(34),0)

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foci 0=-400/9 x^2+16y^2+400

foci

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400

(34, 0), (- 34, 0)

Solution steps

(h + c, k), (h - c, k)

Calculate Hyperbola properties

0 = - \frac{400}{9} x^{2} + 16 y^{2} + 400 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 3, b = 5

(0 + c, 0), (0 - c, 0)

Compute

c :

c = 3^{2} + 5^{2} : 34

(0 + 34, 0), (0 - 34, 0)

Refine

(34, 0), (- 34, 0)

5 x^{2} + 7 y = 3

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

x^{2} + 14 x + y^{2} - 6 y - 23 = 0

x = 2 y - y^{2}

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

What is the foci 0=-400/9 x^2+16y^2+400 ?
The foci 0=-400/9 x^2+16y^2+400 is (sqrt(34),0),(-sqrt(34),0)