What is the vertices x^2-9y^2-4x+36y-41=0 ?

The vertices x^2-9y^2-4x+36y-41=0 is (5,2),(-1,2)

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vertices x^2-9y^2-4x+36y-41=0

vertices

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0

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

Solution steps

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

Calculate Hyperbola properties

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0 :

Right-left Hyperbola with

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

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

Refine

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

\frac{x ^{2}}{9} - \frac{y ^{2}}{49} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 1 ) ^{2}}{49} = 1

\frac{y ^{2}}{9} - \frac{x ^{2}}{4} = 1

\frac{x ^{2}}{100} - \frac{y ^{2}}{64} = 1

What is the vertices x^2-9y^2-4x+36y-41=0 ?
The vertices x^2-9y^2-4x+36y-41=0 is (5,2),(-1,2)