What is the vertices ((x-3)^2)/(16)-((y-2)^2)/(25)=1 ?

The vertices ((x-3)^2)/(16)-((y-2)^2)/(25)=1 is (7,2),(-1,2)

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vertices ((x-3)^2)/(16)-((y-2)^2)/(25)=1

vertices

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

(7, 2), (- 1, 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (3, 2), a = 4, b = 5

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

Refine

(7, 2), (- 1, 2)

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

9 y^{2} - 16 x^{2} = 144

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

9 x^{2} - 25 y^{2} - 72 x + 369 = 0

\frac{( x + 2 ) ^{2}}{20} - \frac{y ^{2}}{20} = 1

What is the vertices ((x-3)^2)/(16)-((y-2)^2)/(25)=1 ?
The vertices ((x-3)^2)/(16)-((y-2)^2)/(25)=1 is (7,2),(-1,2)