Answer:
Step-by-step explanation:
Given
[tex]y^2=1-3x^2[/tex]
[tex]3x^2+y^2=1[/tex]
[tex]\frac{x^2}{(\frac{1}{\sqrt{3}})^2}+\frac{y^2}{1}=1[/tex]
therefore it is a vertical ellipse
thus a=1
[tex]b=\frac{1}{\sqrt{3}}[/tex]
eccentricity of Ellipse
[tex]e^2=1-\frac{b^2}{a^2}[/tex]
[tex]e^2=1-\frac{1}{(\sqrt{3})^2}[/tex]
[tex]e^2=1-\frac{1}{3}[/tex]
[tex]e^2=\frac{2}{3}[/tex]
[tex]e=\sqrt{\frac{2}{3}}[/tex]
Focii are [tex](0,ae)[/tex] and [tex](0,-ae)[/tex]
[tex]ae=1\times \sqrt{\frac{2}{3}}[/tex]
thus focii are [tex](0,\sqrt{\frac{2}{3}})[/tex] & [tex](0,-\sqrt{\frac{2}{3}})[/tex]
(b) Length of major axis [tex]=2a=2\times 1[/tex]
length of minor axis[tex]=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}[/tex]
