Chapter 10.5, Problem 57E

To determine

Find any points of intersection of the graphs of the equations algebraically and then verify using a graphing utility.

Answer to Problem 57E

The points of intersections of the given conics are:

  $\left(1,0\right)$ .

Explanation of Solution

Given:

  $\begin{array}{l}{x}^2+4y^2–2x–8y+1=\mathrm{0......}\left(1\right)\hfill \\ –x^2+2x–4y–1=\mathrm{0.....}\left(2\right)\hfill \end{array}$

Consider the following two equations of the conic,

  $\begin{array}{l}{x}^2+4y^2–2x–8y+1=\mathrm{0......}\left(1\right)\hfill \\ –x^2+2x–4y–1=\mathrm{0.....}\left(2\right)\hfill \end{array}$

Adding the above two equations,

  $\begin{array}{l}\begin{array}{l}{x}^2+4y^2–2x–8y+1+\left(–x^2+2x–4y–1\right)=0\hfill \\ 4y^2–12y=0\hfill \\ y^2–3y=0\hfill \\ y\left(y–3\right)=0\hfill \end{array}\\ y=0,3\end{array}$

For the value of $y=0$ the value of x is as below,

  $\begin{array}{l}–x^2+2x–4\times 0–1=0\hfill \\ x^2–2x+1=0\hfill \\ {\left(x–1\right)}^2=0\hfill \\ x=1\hfill \end{array}$

For thevalue of $y=3$ the value of x is as below,

  $\begin{array}{l}–x^2+2x–4\times 3–1=0\hfill \\ x^2–2x+13=0\hfill \\ \begin{array}{l}x=\frac{2\pm \sqrt{4^2-4\times 13}}{2}\hfill \\ x=\frac{2\pm \sqrt{-36}}{2}\hfill \end{array}\hfill \\ \hfill \end{array}$

Then the value of x is not real.

Therefore the point of intersections of the given conics is:

  $\left(1,0\right)$

The sketch of the graphs is given below using Maple.

Therefore the points of intersection can be verified by moving the cursor over the points of intersection in the graph and the graph is as above.