Question

An equation of an ellipse is

given. (a) Find the vertices, foci, and eccentricity of the ellipse.

(b) Determine the lengths of the major and minor axes. (c) Sketch

a graph of the ellipse.

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

Short answer

a. Hence, for the equation $x^2+4y^2=1$,the vertices are$(\pm 1,0)$,the foci are$(\pm \frac{\sqrt{3}}{2},0)$and the eccentricity is$\frac{\sqrt{3}}{2}$.

b. Hence, for the equation $x^2+4y^2=1$,the vertical length of the major axis is 4 units and the horizontal length of the minor axis is1units.

c.

Step-by-step solution

a.Step 1. Given information.

Given the equation of ellipse $x^2+4y^2=1$.

$\Rightarrow \frac{x^2}{1}+\frac{y^2}{1/4}=1$

Step 2. Write down the concept.

For foci, use the formula $c^2=a^2-b^2$

Here, a = 1

$b=\frac{1}{2}$

So, $c^2=1^2-\frac{1}{4}$

$\begin{array}{l}\Rightarrow c^2=\frac{3}{4}\\ \Rightarrow c=\frac{\pm \sqrt{3}}{2}\end{array}$

The foci are $\left(\pm \frac{\sqrt{3}}{2},0\right)$

The eccentricity is given by $e=\frac{c}{a}$

$\Rightarrow e=\frac{\sqrt{3}}{2}$

b.Step 1. Given information.

Given the equation of ellipse $x^2+4y^2=1$.

Step 2. Write down the concept.

Given $x^2+3y^2=9$

$\Rightarrow \frac{x^2}{1}+\frac{y^2}{1/4}=1$

The vertical length of the major axis is 2a= $2\times 1$

$\Rightarrow 2a=2$

The horizontal length of the minor axis is 2b = $2\times \frac{1}{2}$ $\Rightarrow 2b=1$

c.Step 1. Given information.

Given the equation of ellipse $x^2+4y^2=1$.

Step 2. Write down the concept.

The equation of ellipse $x^2+4y^2=1$.

$\Rightarrow \frac{x^2}{1}+\frac{y^2}{1/4}=1$

On comparing the above with the standard equation $\frac{{(x-h)}^2}{b^2}+\frac{{(y-k)}^2}{a^2}=1$, we get

The center (h, k) at (0, 0)

$a=1$

$b=\frac{1}{4}$

The points are $(1,0),(-1,0),(0,\frac{1}{2})$ and $(0,-\frac{1}{2})$.