What is the vertices ((x-3)^2)/4-(y^2)/(25)=1 ?

The vertices ((x-3)^2)/4-(y^2)/(25)=1 is (5,0),(1,0)

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vertices ((x-3)^2)/4-(y^2)/(25)=1

vertices

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

(5, 0), (1, 0)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (3, 0), a = 2, b = 5

(3 + 2, 0), (3 - 2, 0)

Refine

(5, 0), (1, 0)

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

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

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

x^{2} + 5 x - y^{2} + 3 y = 1

9 x^{2} - 4 y^{2} + 36 x + 8 y + 33 = 0

What is the vertices ((x-3)^2)/4-(y^2)/(25)=1 ?
The vertices ((x-3)^2)/4-(y^2)/(25)=1 is (5,0),(1,0)