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Popular >

vertices-4x^2+y^2=16

Solution

vertices

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

Solution

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 2

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

Refine

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

Popular Examples

vertices x^2-y^2+6x-4y-4=0 vertices

x^{2} - y^{2} + 6 x - 4 y - 4 = 0

vertices 16x^2-9y^2-64x-18y+199=0 vertices

16 x^{2} - 9 y^{2} - 64 x - 18 y + 199 = 0

vertices (x^2)/(36)-(y^2)/(81)=1 vertices

\frac{x ^{2}}{36} - \frac{y ^{2}}{81} = 1

vertices ((y-1)^2)/9-((x-2)^2)/(16)=1 vertices

\frac{( y - 1 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

vertices ((x+1)^2)/(25)-((y+4)^2)/(64)=1 vertices

\frac{( x + 1 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{64} = 1

Frequently Asked Questions (FAQ)

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