What is the vertices (x^2)/7-(7y^2)/(42)=1 ?

The vertices (x^2)/7-(7y^2)/(42)=1 is (sqrt(7),0),(-sqrt(7),0)

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vertices (x^2)/7-(7y^2)/(42)=1

vertices

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1

(7, 0), (- 7, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{7} - \frac{7 y ^{2}}{42} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 7, b = 6

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

x^{2} = \frac{9}{16} y^{2} + 9

9 x^{2} - 4 y^{2} - 36 x + 8 y - 4 = 0

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

y^{2} - x^{2} = 1

\frac{( x - 5 ) ^{2}}{81} - \frac{( y - 7 ) ^{2}}{49} = 1

What is the vertices (x^2)/7-(7y^2)/(42)=1 ?
The vertices (x^2)/7-(7y^2)/(42)=1 is (sqrt(7),0),(-sqrt(7),0)