What is the vertices ((x-3)^2)/(49)-((y-2)^2)/(16)=1 ?

The vertices ((x-3)^2)/(49)-((y-2)^2)/(16)=1 is (10,2),(-4,2)

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vertices ((x-3)^2)/(49)-((y-2)^2)/(16)=1

vertices

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 2 ) ^{2}}{16} = 1

(10, 2), (- 4, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{49} - \frac{( y - 2 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (3, 2), a = 7, b = 4

(3 + 7, 2), (3 - 7, 2)

Refine

(10, 2), (- 4, 2)

(\frac{x}{9})^{2} - (\frac{y}{49})^{2} = 1

4 x^{2} - 8 y^{2} = 3

2 x^{2} - 3 y^{2} = 6

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

(\frac{y - 4}{16})^{2} - (\frac{x - 6}{16})^{2} = 1

What is the vertices ((x-3)^2)/(49)-((y-2)^2)/(16)=1 ?
The vertices ((x-3)^2)/(49)-((y-2)^2)/(16)=1 is (10,2),(-4,2)