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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.

User Kshitij
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1 Answer

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Answer:

Explanation:

Given


y^2=1-3x^2


3x^2+y^2=1


(x^2)/(((1)/(√(3)))^2)+(y^2)/(1)=1

therefore it is a vertical ellipse

thus a=1


b=(1)/(√(3))

eccentricity of Ellipse


e^2=1-(b^2)/(a^2)


e^2=1-(1)/((√(3))^2)


e^2=1-(1)/(3)


e^2=(2)/(3)


e=\sqrt{(2)/(3)}

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


ae=1* \sqrt{(2)/(3)}

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

(b) Length of major axis
=2a=2* 1

length of minor axis
=2b=2* \sqrt{(2)/(3)}=2\cdot \sqrt{(2)/(3)}

User Tinisha
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