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Popular >

vertices x^2+16y^2=1

Solution

vertices

x^{2} + 16 y^{2} = 1

Solution

(1, 0), (- 1, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 16 y^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 1, b = \frac{1}{4}

(0 + 1, 0), (0 - 1, 0)

Refine

(1, 0), (- 1, 0)

Popular Examples

vertices 2x^2+7y^2=14 vertices

2 x^{2} + 7 y^{2} = 14

vertices 16x^2+25y^2-1600=0 vertices

16 x^{2} + 25 y^{2} - 1600 = 0

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

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

vertices x^2+10x+9y^2+16=0 vertices

x^{2} + 10 x + 9 y^{2} + 16 = 0

vertices (x^2}{16}+\frac{y^2)/7 =1 vertices

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

Frequently Asked Questions (FAQ)

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