What is the vertices (x^2)/(60)-(y^2)/(40)=1 ?

The vertices (x^2)/(60)-(y^2)/(40)=1 is (2sqrt(15),0),(-2sqrt(15),0)

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vertices (x^2)/(60)-(y^2)/(40)=1

vertices

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

(215, 0), (- 215, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 215, b = 210

(0 + 215, 0), (0 - 215, 0)

Refine

(215, 0), (- 215, 0)

\frac{x ^{2}}{36} - \frac{y ^{2}}{25} = 1

x^{2} - 9 y^{2} = 36

\frac{x ^{2}}{64} - \frac{y ^{2}}{49} = 1

\frac{( y - 5 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 1

9 x^{2} - 4 y^{2} + 54 x + 16 y + 66 = 0

What is the vertices (x^2)/(60)-(y^2)/(40)=1 ?
The vertices (x^2)/(60)-(y^2)/(40)=1 is (2sqrt(15),0),(-2sqrt(15),0)