What is the vertices ((x-3)^2)/3-((y-2)^2)/(22)=1 ?

The vertices ((x-3)^2)/3-((y-2)^2)/(22)=1 is (3+sqrt(3),2),(3-sqrt(3),2)

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vertices ((x-3)^2)/3-((y-2)^2)/(22)=1

vertices

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{3} - \frac{( y - 2 ) ^{2}}{22} = 1 :

Right-left Hyperbola with

(h, k) = (3, 2), a = 3, b = 22

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

16 x^{2} - 9 y^{2} + 64 x - 90 y = 305

4 x^{2} - 9 y^{2} - 36 < 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

What is the vertices ((x-3)^2)/3-((y-2)^2)/(22)=1 ?
The vertices ((x-3)^2)/3-((y-2)^2)/(22)=1 is (3+sqrt(3),2),(3-sqrt(3),2)