Chapter B, Problem 50E

To determine

To calculate: To find the foci of the ellipse $x^2+4y^2=4$ and sketch its graph

Answer to Problem 50E

The foci of ellipse are $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$

Explanation of Solution

Given information: Equation of ellipse is $x^2+4y^2=4$

Formula Used:

Standard equation of ellipse is

  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ,

where $c^2=a^2-b^2$

Foci of above ellipse is $\left(\pm c,0\right)$

Calculation:

Equation of ellipse is given as

  $x^2+4y^2=4$

Rewrite the equation as follows:

  $\begin{array}{l}\frac{x^2}{4}+\frac{4y^2}{4}=\frac{4}{4}\\ \frac{x^2}{4}+\frac{y^2}{1}=1\\ \frac{x^2}{2^2}+\frac{y^2}{1^2}=1\end{array}$

On comparing the equation with standard equation of ellipse,

  $a=2$

  $b=1$

Value of $c$ is calculated as follows:

  $\begin{array}{l}{c}^2=a^2-b^2\\ c^2=2^2-1^2\\ c^2=4-1\\ c^2=3\\ c=\pm \sqrt{3}\end{array}$

Thus, foci of ellipse are $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$

On sketching the ellipse:

  

Conclusion:

Hence, foci of ellipse are $\left(\sqrt{3},0\right)$ and $\left(-\sqrt{3},0\right)$