What is the vertices (x^2)/(16)-(y^2)/(20)=1 ?

The vertices (x^2)/(16)-(y^2)/(20)=1 is (4,0),(-4,0)

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vertices (x^2)/(16)-(y^2)/(20)=1

vertices

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{16} - \frac{y ^{2}}{20} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 4, b = 25

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

Refine

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

x^{2} = 24 y

3 x^{2} + 3 x + 2 y = 0

x^{2} + y^{2} - 2 x + 4 y + 1 = 0

4 x^{2} + y^{2} = 16

y^{2} = x - 2

What is the vertices (x^2)/(16)-(y^2)/(20)=1 ?
The vertices (x^2)/(16)-(y^2)/(20)=1 is (4,0),(-4,0)