Chapter 10.5, Problem 55E

To determine

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

Answer to Problem 55E

The points of intersections of the given conics are:

  $\left(14,–8\right)$ and $\left(6,–8\right)$ .

Explanation of Solution

Given:

Consider the conic equation as:

  $\begin{array}{l}{x}^2–4y^2–20x–64y–172=\mathrm{0......}\left(1\right)\hfill \\ 16x^2+4y^2–320x+64y+1600=\mathrm{0.....}\left(2\right)\hfill \end{array}$

Consider the following two equations of the conic,

  $\begin{array}{l}{x}^2–4y^2–20x–64y–172=\mathrm{0......}\left(1\right)\hfill \\ 16x^2+4y^2–320x+64y+1600=\mathrm{0.....}\left(2\right)\hfill \end{array}$

Add the above two equations as shown below,

  $\begin{array}{l}{x}^2–4y^2–20x–6y–172+\left(16x^2+4y^2–320x+64y+1600\right)=0\hfill \\ 17x^2–340x+1428=0\hfill \\ x^2–20x+84=0\hfill \end{array}$

Solve for the value of x as shown below:

  $\begin{array}{cc}& x=\frac{20\pm \sqrt{{(-20)}^2-4\times 84}}{2}\\ & =\frac{20\pm \sqrt{64}}{2}\\ & =\frac{20\pm 8}{2}\\ & =14,6\end{array}$

For the value of $x=14$ the value of y is as below,

  $\begin{array}{l}{14}^2–4y^2–20\times 14–64y–172=0\hfill \\ 196–280–172–4y^2–64y=0\hfill \\ –256–4y^2–64y=0\hfill \end{array}$

Further solve as below:

  $\begin{array}{l}{y}^2+16y+64=0\hfill \\ {\left(y+8\right)}^2=0\hfill \\ y=–8\hfill \end{array}$

For the value of $x=6$ the value of y is as below,

  $\begin{array}{l}{6}^2–4y^2–20\times 6–64y–172=0\hfill \\ 36–120–172–4y^2–64y=0\hfill \\ –256–4y^2–64y=0\hfill \end{array}$

Further solve as below:

  $\begin{array}{l}{y}^2+16y+64=0\hfill \\ {\left(y+8\right)}^2=0\hfill \\ y=–8\hfill \end{array}$

Therefore the points of intersections of the given conics are:

  $\left(14,–8\right)$ and $\left(6,–8\right)$

The sketch of the graphs is given 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.