What is the vertices ((x-6)^2)/(48)-(y-2)^2=1 ?

The vertices ((x-6)^2)/(48)-(y-2)^2=1 is (6+4sqrt(3),2),(6-4sqrt(3),2)

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vertices ((x-6)^2)/(48)-(y-2)^2=1

vertices

\frac{( x - 6 ) ^{2}}{48} - (y - 2)^{2} = 1

(6 + 43, 2), (6 - 43, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 6 ) ^{2}}{48} - (y - 2)^{2} = 1 :

Right-left Hyperbola with

(h, k) = (6, 2), a = 43, b = 1

(6 + 43, 2), (6 - 43, 2)

x^{2} - \frac{y ^{2}}{25} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

y^{2} - 25 x^{2} + 8 y - 9 = 0

x^{2} - 16 y^{2} = 144

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

What is the vertices ((x-6)^2)/(48)-(y-2)^2=1 ?
The vertices ((x-6)^2)/(48)-(y-2)^2=1 is (6+4sqrt(3),2),(6-4sqrt(3),2)