What is the vertices ((x-3)^2)/(49)-((y-5)^2)/(15)=1 ?

The vertices ((x-3)^2)/(49)-((y-5)^2)/(15)=1 is (10,5),(-4,5)

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vertices ((x-3)^2)/(49)-((y-5)^2)/(15)=1

vertices

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 5 ) ^{2}}{15} = 1

(10, 5), (- 4, 5)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 5 ) ^{2}}{15} = 1 :

Right-left Hyperbola with

(h, k) = (3, 5), a = 7, b = 15

(3 + 7, 5), (3 - 7, 5)

Refine

(10, 5), (- 4, 5)

\frac{( x - 6 ) ^{2}}{8} - \frac{( y - 5 ) ^{2}}{1} = 1

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

\frac{x ^{2} - y ^{2}}{24} = 1

x^{2} + 6 x = 9 y^{2}

\frac{y ^{2}}{4 8 ^{2}} - \frac{x ^{2}}{2 0 ^{2}} = 1

What is the vertices ((x-3)^2)/(49)-((y-5)^2)/(15)=1 ?
The vertices ((x-3)^2)/(49)-((y-5)^2)/(15)=1 is (10,5),(-4,5)