What is the vertices ((x-1)^2)/9-((y-2)^2)/(16)=1 ?

The vertices ((x-1)^2)/9-((y-2)^2)/(16)=1 is (4,2),(-2,2)

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vertices ((x-1)^2)/9-((y-2)^2)/(16)=1

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (1, 2), a = 3, b = 4

(1 + 3, 2), (1 - 3, 2)

Refine

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

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

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

\frac{( x - 3 ) ^{2}}{16} - \frac{( y + 4 ) ^{2}}{25} = 1

5 x^{2} - 5 y^{2} + 20 x - 60 y = 185

9 x^{2} - 4 y^{2} + 54 x - 8 y + 78 = 0

What is the vertices ((x-1)^2)/9-((y-2)^2)/(16)=1 ?
The vertices ((x-1)^2)/9-((y-2)^2)/(16)=1 is (4,2),(-2,2)