What is the vertices ((x+3)^2)/(49)-((y-1)^2)/(48)=1 ?

The vertices ((x+3)^2)/(49)-((y-1)^2)/(48)=1 is (4,1),(-10,1)

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vertices ((x+3)^2)/(49)-((y-1)^2)/(48)=1

vertices

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1

(4, 1), (- 10, 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1 :

Right-left Hyperbola with

(h, k) = (- 3, 1), a = 7, b = 43

(- 3 + 7, 1), (- 3 - 7, 1)

Refine

(4, 1), (- 10, 1)

\frac{9}{125} y^{2} - \frac{1}{5} x^{2} = 1

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

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

x^{2} - y^{2} + 2 y = 2

9 y^{2} + 36 y - 4 x^{2} - 32 x - 64 = 0

What is the vertices ((x+3)^2)/(49)-((y-1)^2)/(48)=1 ?
The vertices ((x+3)^2)/(49)-((y-1)^2)/(48)=1 is (4,1),(-10,1)