What is the vertices (x^2)/(49)-(y^2)/(16)=1 ?

The vertices (x^2)/(49)-(y^2)/(16)=1 is (7,0),(-7,0)

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vertices (x^2)/(49)-(y^2)/(16)=1

vertices

\frac{x ^{2}}{49} - \frac{y ^{2}}{16} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{49} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 7, b = 4

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

Refine

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{9} = 1

x = - \frac{1}{8} y^{2}

y^{2} = - 32 x

x^{2} - 6 x + 5 y = - 34

x = - 3 y^{2}

What is the vertices (x^2)/(49)-(y^2)/(16)=1 ?
The vertices (x^2)/(49)-(y^2)/(16)=1 is (7,0),(-7,0)