Chapter 11, Problem 35RE

To determine

To find: The type of curve, find the focus, vertices and to plot the equation in a graph.

Answer to Problem 35RE

The given type of curve is an ellipse with:

Co-ordinates of vertices $\left(1,4\pm \sqrt{20}\right)$

Co-ordinates of focus $\left(1,4\pm \sqrt{15}\right)$

Explanation of Solution

Given information:

The equation of the curve is:

  $4x^2+y^2=8(x+y)$

Concept used:

This is the equation of ellipse whose center is not at origin $(0,0)$ . Standard equation of ellipse when center is $(h,k)$ is given by:

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

If $a^2>b^2$ ,

Co-ordinates of vertices $\left(h\pm a,k\right)$

Co-ordinates of foci $\left(h\pm c,k\right)$

If $a^2<b^2$ ,

Co-ordinates of vertices $\left(h,k\pm b\right)$

Co-ordinates of foci $\left(h,k\pm c\right)$

Calculation:

  $\begin{array}{l}4x^2+y^2-8x-8y=0\\ (4x^2-8x)+(y^2-8y)=0\\ (4x^2-8x+4-4)+(y^2-8y+16-16)=0\\ (4x^2-8x+4)+(y^2-8y+16)-20=0\end{array}$

  $\begin{array}{l}4{(}^x+{(}^y-20=0\\ 4{(}^x+{(}^y=20\\ \frac{{(x-1)}^2}{5}+\frac{{(y-4)}^2}{20}=1\to 2\end{array}$

Here, $20>5$ , Major axis is $y-axis.$

  $\begin{array}{l}{b}^2=20\Rightarrow b=\sqrt{20}\\ a^2=5\Rightarrow a=\sqrt{5}\\ c^2=15\Rightarrow c=\sqrt{15}\end{array}$

Co-ordinates of vertices $\left(1,4\pm \sqrt{20}\right)$

Co-ordinates of focus $\left(1,4\pm \sqrt{15}\right)$

Graph:

  

Interpretation :

The given type of curve is an ellipse with:

Co-ordinates of vertices $\left(1,4\pm \sqrt{20}\right)$

Co-ordinates of focus $\left(1,4\pm \sqrt{15}\right)$