What is the vertices (x^2)/(16)-(y^2)/(36)=1 ?

The vertices (x^2)/(16)-(y^2)/(36)=1 is (4,0),(-4,0)

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vertices (x^2)/(16)-(y^2)/(36)=1

vertices

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{36} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 6

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

Refine

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

(x + 2)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{4}} = 1

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x = - 5 y^{2}

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

What is the vertices (x^2)/(16)-(y^2)/(36)=1 ?
The vertices (x^2)/(16)-(y^2)/(36)=1 is (4,0),(-4,0)