Chapter 8.6, Problem 38E

Solution

To find: The type of conic section, write equation in standard form and graph of the equation.

The equation $x^2+4y^2-10x+16y+37=0$ is an ellipse and the standard form is $\frac{{\left(x-5\right)}^2}{4}+\frac{{\left(y+2\right)}^2}{1}=1$ . The graph is shown in figure (1).

Given information: The given equation is $x^2+4y^2-10x+16y+37=0$.

Formula used: The general second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ describe any conic.

The type of conic is a circle if $B^2-4AC<0,B=0,A=C$.

The type of conic is an ellipse if $B^2-4AC<0,B\ne 0,A\ne C$.

The type of conic is a parabola if $B^2-4AC=0$.

The type of conic is a hyperbola if $B^2-4AC>0$.

Calculation:

Consider the given equation.

  $x^2+4y^2-10x+16y+37=0$

Compare the above equation with the general equation of the conic.

  $A=1,B=0,C=4$

Substitute $1$ for $A$ , $0$ for $B$ and $4$ for $C$ in $B^2-4AC$ to find the discriminant.

  $\begin{array}{c}{B}^2-4AC={\left(0\right)}^2-4\left(4\right)\left(1\right)\\ =0-16\\ =-16\end{array}$

Since $-16<0$ and $A\ne C$ . So, the equation satisfies all the conditions of an ellipse.

Rewrite the given equation in the standard form to complete the square.

  $\begin{array}{c}{x}^2+4y^2-10x+16y=-37\\ \left(x^2-10x+25\right)-25+4\left(y^2+4y+4\right)-16=-37\\ {\left(x-5\right)}^2+4{\left(y+2\right)}^2=4\\ \frac{{\left(x-5\right)}^2}{4}+\frac{{\left(y+2\right)}^2}{1}=1\end{array}$

The center of the ellipse is $\left(5,-2\right)$ , $a=2$ and $b=1$.

Drawn the graph of the ellipse using the above conditions.

  

Figure (1)

Therefore, the equation $x^2+4y^2-10x+16y+37=0$ is an ellipse and the standard form is $\frac{{\left(x-5\right)}^2}{4}+\frac{{\left(y+2\right)}^2}{1}=1$.