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)…

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asked Oct 15, 2020 228k views

1 Answer

Tinisha · Answer 1 · 2020-10-20T01:16:42+0000

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
$=2b=2* \sqrt{(2)/(3)}=2\cdot \sqrt{(2)/(3)}$