Question

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.

Answer 1

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$=2b=2\times \sqrt{\frac{2}{3}}=2\cdot \sqrt{\frac{2}{3}}$