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UkoKoshka [18]
3 years ago
6

An equation of an ellipse is given. y2 = 1 − 3x2 (a) Find the vertices, foci, and eccentricity of the ellipse. vertex (x, y) = (

smaller y-value) vertex (x, y) = (larger y-value) focus (x, y) = (smaller y-value) focus (x, y) = (larger y-value) eccentricity (b) Determine the length of the major axis. Determine the length of the minor axis.
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
8 0

Answer:

Step-by-step explanation:

Given

y^2=1-3x^2

3x^2+y^2=1

\frac{x^2}{(\frac{1}{\sqrt{3}})^2}+\frac{y^2}{1}=1

therefore it is a vertical ellipse

thus a=1

b=\frac{1}{\sqrt{3}}

eccentricity of Ellipse

e^2=1-\frac{b^2}{a^2}

e^2=1-\frac{1}{(\sqrt{3})^2}

e^2=1-\frac{1}{3}

e^2=\frac{2}{3}

e=\sqrt{\frac{2}{3}}

Focii are (0,ae) and (0,-ae)

ae=1\times \sqrt{\frac{2}{3}}

thus focii are (0,\sqrt{\frac{2}{3}}) & (0,-\sqrt{\frac{2}{3}})

(b) Length of major axis =2a=2\times 1

length of minor axis=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}

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