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

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

\frac{x ^{2}}{4} - \frac{( y + 3 ) ^{2}}{9} = 1

25 y^{2} - 16 x^{2} - 400 = 0

x^{2} - 9 y^{2} + 4 x + 18 y - 149 = 0

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)