Chapter 8.4, Problem 4E

Solution

To calculate: The vertices and foci of the ellipse $5x^2+6xy+5y^2=16$ by analyzing the graph.

The vertices of the ellipse $5x^2+6xy+5y^2=16$ are $\left(-2,2\right)$ and $\left(2,-2\right)$ , and the foci are $\left(-\sqrt{3},\sqrt{3}\right)$ and $\left(\sqrt{3},-\sqrt{3}\right)$ .

Given information: The given equation of the ellipse is $5x^2+6xy+5y^2=16$ . The graph is symmetric about the lines $y=x$ and $y=-x$ .

Calculation:

Draw the graph of the ellipse $5x^2+6xy+5y^2=16$ and lines of symmetry as shown below.

  

Figure (1)

From the graph, it can be observed that the vertices of the ellipse are $\left(-2,2\right)$ and $\left(2,-2\right)$ .

The foci lie on the line $y=x$ and $c$ units from the origin. The distance of a vertex to the center is $a$ . Consider the vertex $\left(-2,2\right)$ .

Calculate the value of $a$ .

  $\begin{array}{c}a=\sqrt{{\left(-2\right)}^2+2^2}\\ =\sqrt{4+4}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

The co-vertices of the ellipse are $\left(-1,-1\right)$ and $\left(1,1\right)$ . The distance of a co-vertex to the center is $b$ . Consider the vertex $\left(1,1\right)$ .

Calculate the value of $b$ .

  $\begin{array}{c}b=\sqrt{1^2+1^2}\\ =\sqrt{1+1}\\ =\sqrt{2}\end{array}$

Calculate the value of $c$ by the formula $c^2=a^2-b^2$ .

  $\begin{array}{c}c=\sqrt{{\left(2\sqrt{2}\right)}^2-{\left(\sqrt{2}\right)}^2}\\ =\sqrt{8-2}\\ =\sqrt{6}\end{array}$

So, the foci is at $\sqrt{6}$ units distance from the center along the line $y=x$ . Let the foci be $\left(-m,m\right)$ and $\left(m,-m\right)$ .

Apply Pythagorean theorem and find the value of $m$ .

  $\begin{array}{c}{m}^2+m^2={\left(\sqrt{6}\right)}^2\\ 2m^2=6\\ m=\sqrt{3}\end{array}$

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

Thus, the vertices of the ellipse $5x^2+6xy+5y^2=16$ are $\left(-2,2\right)$ and $\left(2,-2\right)$ , and the foci are $\left(-\sqrt{3},\sqrt{3}\right)$ and $\left(\sqrt{3},-\sqrt{3}\right)$ .