What is the vertices 5y^2-16x^2+400=0 ?

The vertices 5y^2-16x^2+400=0 is (5,0),(-5,0)

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vertices 5y^2-16x^2+400=0

vertices

5 y^{2} - 16 x^{2} + 400 = 0

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

5 y^{2} - 16 x^{2} + 400 = 0 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 45

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

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

\frac{( y ) ^{2}}{40} - \frac{( x ) ^{2}}{81} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1

6 x^{2} - 8 y^{2} + 12 x - 32 y = 178

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

What is the vertices 5y^2-16x^2+400=0 ?
The vertices 5y^2-16x^2+400=0 is (5,0),(-5,0)