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

The vertices x^2-y^2-4x+4y+9=0 is (2,5),(2,-1)

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

vertices

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

(2, 5), (2, - 1)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

(2, 2 + 3), (2, 2 - 3)

Refine

(2, 5), (2, - 1)

x^{2} - y^{2} + 8 x - 2 y - 10 = 0

25 y^{2} - 9 x^{2} = 225

9 x^{2} - y^{2} = 36

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

\frac{x ^{2}}{4} - \frac{y ^{2}}{36} = 1

What is the vertices x^2-y^2-4x+4y+9=0 ?
The vertices x^2-y^2-4x+4y+9=0 is (2,5),(2,-1)