Chapter 8.6, Problem 14GP

Solution

To classify the path of the asteroid. Then find the standard equation and graph it.

The path of the asteroid is an ellipse whose standard equation is $\frac{{\left(x-\frac{3}{2}\right)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{y^2}{2^2}=1$ , and the graph is shown below.

  

Given information:

The conic, $4x^2+6.25y^2-12x-16=0$.

Calculation:

Rewrite the equation $4x^2+6.25y^2-12x-16=0$ in the form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$.

  $\begin{array}{c}4x^2+6.25y^2-12x-16=0\\ 4x^2+0\cdot xy+6.25y^2-12x+0\cdot y-16=0\end{array}$

Note that for the equation $4x^2+6.25y^2-12x-16=0$ , $A=4$ , $B=0$ and $C=6.25$ . So, the value of the discriminant $B^2-4AC$ is:

  $\begin{array}{c}{B}^2-4AC=0^2-4\cdot 4\cdot 6.25\\ =-100\end{array}$

Since $B^2-4AC<0$ and $A\ne C$ , the conic is an ellipse.

Now, rewrite the conic in standard form.

  $\begin{array}{c}4x^2+6.25y^2-12x-16=0\\ 4x^2-12x+6.25y^2=16\\ 4\left(x^2-3x\right)+6.25y^2=16\\ 4\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)+6.25y^2=16\\ 4\left(x^2-3x+\frac{9}{4}\right)-9+6.25y^2=16\\ 4{\left(x-\frac{3}{2}\right)}^2+6.25y^2=25\\ \frac{{\left(x-\frac{3}{2}\right)}^2}{\left(\frac{25}{4}\right)}+\frac{y^2}{\left(\frac{25}{6.25}\right)}=1\\ \frac{{\left(x-\frac{3}{2}\right)}^2}{\left(\frac{25}{4}\right)}+\frac{y^2}{4}=1\\ \frac{{\left(x-\frac{3}{2}\right)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{y^2}{2^2}=1\end{array}$

Therefore, the standard equation of the conic is $\frac{{\left(x-\frac{3}{2}\right)}^2}{{\left(\frac{5}{2}\right)}^2}+\frac{y^2}{2^2}=1$.

The ellipse has the center at $\left(\frac{3}{2},0\right)$ with major axis of length

From the equation, the ellipse has center at $\left(\frac{3}{2},0\right)$ , $a=\frac{5}{2}$ and $b=2$ . The vertices of the ellipse are at $\left(\frac{3}{2}\pm a,0\right)$ or $\left(-1,0\right)$ and $\left(4,0\right)$ , and covertices are at $\left(\frac{3}{2},0\pm b\right)$ or $\left(\frac{3}{2},-2\right)$ and $\left(\frac{3}{2},2\right)$.

Plot the center, vertices and covertices and then draw the ellipse.