What is the vertices 9(x-3)^2-4(y-1)^2=-36 ?

The vertices 9(x-3)^2-4(y-1)^2=-36 is (3,4),(3,-2)

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vertices 9(x-3)^2-4(y-1)^2=-36

vertices

9 (x - 3)^{2} - 4 (y - 1)^{2} = - 36

(3, 4), (3, - 2)

Solution steps

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

Calculate Hyperbola properties

9 (x - 3)^{2} - 4 (y - 1)^{2} = - 36 :

Up-down Hyperbola with

(h, k) = (3, 1), a = 3, b = 2

(3, 1 + 3), (3, 1 - 3)

Refine

(3, 4), (3, - 2)

\frac{y ^{2}}{144} - \frac{x ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{16} = 1

\frac{x ^{2}}{27} - \frac{y ^{2}}{216} = 1

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

- 9 x^{2} + 16 y^{2} - 36 x - 160 y + 220 = 0

What is the vertices 9(x-3)^2-4(y-1)^2=-36 ?
The vertices 9(x-3)^2-4(y-1)^2=-36 is (3,4),(3,-2)