What is the vertices (y^2)/(4^2)-(x^2)/(3^2)=1 ?

The vertices (y^2)/(4^2)-(x^2)/(3^2)=1 is (0,4),(0,-4)

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vertices (y^2)/(4^2)-(x^2)/(3^2)=1

vertices

\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 3

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

Refine

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

\frac{( y + 2 ) ^{2}}{4} - \frac{( x - 1 ) ^{2}}{45} = 1

\frac{x ^{2}}{729} - \frac{y ^{2}}{484} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1

\frac{9}{125} y^{2} - \frac{1}{5} x^{2} = 1

What is the vertices (y^2)/(4^2)-(x^2)/(3^2)=1 ?
The vertices (y^2)/(4^2)-(x^2)/(3^2)=1 is (0,4),(0,-4)