Compute some properties of an ellipse with a given Cartesian equation

Find the coordinates of the center, the foci, and the vertices, sketch the curve, and determine the eccentricity of the ellipse given by the equation

$4y^2 + 3x^2 = 1.$

First, we have

$4y^2 + 3x^2 = 1 \quad \implies \quad \frac{x^2}{\left( \frac{1}{\sqrt{3}} \right)^2} + \frac{y^2}{ \left( \frac{1}{2} \right)^2} = 1.$

So, this is the equation for an ellipse in standard form with center at $(0,0)$ and with $a = \frac{1}{\sqrt{3}}$ and $b = \frac{1}{2}$ . Therefore,

$c = \sqrt{a^2 - b^2} = \sqrt{\frac{1}{3} - \frac{1}{4}} = \sqrt{\frac{1}{12}} = \frac{\sqrt{3}}{6}.$

So, the foci are

$\left( \frac{\sqrt{3}}{6}, 0\right) \qquad \text{and} \qquad \left( -\frac{\sqrt{3}}{6}, 0 \right).$

Then, the vertices are $pm aN$ , which are

$\left( \frac{\sqrt{3}}{3}, 0 \right), \qquad \text{and} \qquad \left( - \frac{\sqrt{3}}{3}, 0 \right).$

Finally, we compute the eccentricity,

$c = ea \quad \implies \quad \frac{\sqrt{3}}{6} = \frac{1}{\sqrt{3}} e \quad \implies \quad e = \frac{1}{2}.$

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