How do you graph x^2+9y^2=9?

Question

Isiah

Mathematics

21 April, 09:38

How do you graph x^2+9y^2=9?

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Jaylee Chang · Answer 1

x 2 9 - y 2 1 = 1 x29-⁢y21=1 This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola. (x - h) 2 a 2 - (y - k) 2 b 2 = 1 x-⁢h2a2-⁢y-⁢k2b2=1 Match the values in this hyperbola to those of the standard form. The variable h h represents the x-offset from the origin, k k represents the y-offset from origin, a a. a = 3 a=3 b = 1 b=1 k = 0 k=0 h = 0 h=0 The center of a hyperbola follows the form of (h, k) h, k. Substitute in the values of h h and k k. (0, 0) 0,0 Find c c, the distance from the center to a focus.

√ 10 10 Find the vertices.

(3, 0), ( - 3, 0) 3,0,-3,0 Find the foci ... (√ 10, 0), ( - √ 10, 0) 10,0,-⁢10,0 Find the eccentricity.

√ 10 3 103 Find the focal parameter.

√ 10 10 1010 The asymptotes follow the form y = ± b (x - h) a + k y=±b⁢x-⁢ha+k because this hyperbola opens left and right. y = ± 1 3 x + 0 y=±13⁢x+0 Simplify to find the first asymptote.

y = x 3 y=x3 Simplify to find the second asymptote.

y = - x 3 y=-⁢x3 This hyperbola has two asymptotes. y = x 3, y = - x 3 y=x3, y=-⁢x3 These values represent the important values for graphing and analyzing a hyperbola. Center: (0, 0) 0,0 Vertices: (3, 0), ( - 3, 0) 3,0,-3,0 Foci: (√ 10, 0), ( - √ 10, 0) 10,0,-⁢10,0 Eccentricity: √ 10 3 103 Focal Parameter: √ 10 10 1010 Asymptotes : y = x 3 y=x3, y = - x 3 y=-⁢x3