Chapter G, Problem 1PS

To determine

Draw a graph of the ellipse using given properties.

Answer to Problem 1PS

The equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{11}=1$ .

Explanation of Solution

Given:

The distance between two foci is $10$ units and the curve with the sum of the distances from the two foci $12$ units.

Calculation:

A ellipse is the set of all points in a plane such that, for each point on the ellipse, the sum of its distances from two fixed points is a constant.

From definition,

If $\left(x,y\right)$ is any point on the curve and the two fixed points (foci) are $\left(\mp c,0\right)$ , then the difference of $\left(x,y\right)$ from the two fixed points $\left(\mp c,0\right)$ is constant $\left(2a\right)$ .

  $\begin{array}{l}\sqrt{{\left(x+c\right)}^2+{\left(y-0\right)}^2}+\sqrt{{\left(x-c\right)}^2+{\left(y-0\right)}^2}=2a\\ \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\left[b^2=a^2-c^2\right]\end{array}$

Now,

The distance between two foci $=2c$

From question,

  $\begin{array}{l}2c=10\\ \Rightarrow \frac{2c}{2}=\frac{10}{2}\\ \Rightarrow c=5\end{array}$

And

  $\begin{array}{l}2a=12\\ \Rightarrow \frac{2a}{2}=\frac{12}{2}\\ \Rightarrow a=6\end{array}$

  $\begin{array}{l}{b}^2=a^2-c^2\\ b^2=6^2-5^2\\ \Rightarrow b^2=36-25\\ \Rightarrow b^2=11\end{array}$

The equation of ellipse is

  $\begin{array}{l}\frac{x^2}{a^2}+\frac{y^2}{a^2-c^2}=1\\ \Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \Rightarrow \frac{x^2}{36}+\frac{y^2}{11}=1\end{array}$

The standard form of a ellipse is

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$

  $\left(h,k\right)$ is the center of the ellipse.

Rewrite the given equation.

  $\begin{array}{l}\frac{{\left(x-0\right)}^2}{36}+\frac{{\left(y-0\right)}^2}{11}=1\\ \Rightarrow \frac{{\left(x-0\right)}^2}{{\left(6\right)}^2}+\frac{{\left(y-0\right)}^2}{{\left(\sqrt{11}\right)}^2}=1\end{array}$

  $\begin{array}{l}a=6\\ b=\sqrt{11}\\ h=0\\ k=0\end{array}$

Center of the ellipse $\left(h,k\right)=\left(0,0\right)$ .

The distance from the center to a focus.

  $c=5$

Vertex $v_1=\left(h+a,k\right)=\left(0+6,0\right)=\left(6,0\right)\text{and}{v}_2=\left(h-a,k\right)=\left(0-6,0\right)=\left(-6,0\right)$

Focus $f_1=\left(\mp 5,0\right)$

Eccentricity

  $e=\frac{c}{a}=\frac{5}{6}$

The graph of the curve is given below.

  

Hence the equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{11}=1$ .