What is the vertices (x^2)/(16)-(y^2)/(48)=1 ?

The vertices (x^2)/(16)-(y^2)/(48)=1 is (4,0),(-4,0)

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vertices (x^2)/(16)-(y^2)/(48)=1

vertices

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 43

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

x^{2} - 9 y^{2} = 144

4 y^{2} - x^{2} + 2 x + 16 y + 7 = 0

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

- 16 x^{2} + 9 y^{2} = 144

x^{2} - y^{2} + 4 x - 6 y = 6

What is the vertices (x^2)/(16)-(y^2)/(48)=1 ?
The vertices (x^2)/(16)-(y^2)/(48)=1 is (4,0),(-4,0)