91Ó°ÊÓ

Consider the equation.

$6x^2+4y^2+12x-32y+34=0$

Group the $x$terms and width="10">yterms.

localid="1645705118596" $(6x^2+12x)+(4y^2-32y)=-34$

Make the coefficients of $x^2$and $y^2$equal $1$.

$6(x^2+2x)+4(y^2-8y)=-34$

Complete the squares.

$\begin{array}{rcl}6(x^2+2x\mathbf{+}\mathbf{1})+4(y^2-8y\mathbf{+}\mathbf{16})& =& -34\mathbf{+}\mathbf{6}\mathbf{+}\mathbf{64}\\ 6{(x+1)}^2+4{(y-4)}^2& =& -34+70\\ 6{(x+1)}^2+4{(y-4)}^2& =& 36\end{array}$

$\begin{array}{rcl}\frac{6{(x+1)}^2}{36}+\frac{4{(y-4)}^2}{36}& =& \frac{36}{36}\\ \frac{{(x+1)}^2}{6}+\frac{{(y-4)}^2}{9}& =& 1\end{array}$

The standard form of an ellipse with center $(h,k)$is given by localid="1645708013863" $\begin{array}{rcl}\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}& =& 1…\left(1\right)\end{array}$.

Compare the given equation with standard form and conclude that localid="1645708019980" $\begin{array}{rcl}\frac{{(x+1)}^2}{6}+\frac{{(y-4)}^2}{9}& =& 1\end{array}$is in standard form and the center is $(-1,4)$.

Compare the equation $\left(1\right)$with $\begin{array}{rcl}\frac{{(x+1)}^2}{6}+\frac{{(y-4)}^2}{9}& =& 1\end{array}$to find $a^2$and $b^2$.

$\begin{array}{rcl}{a}^2& =& 6\\ b^2& =& 9\end{array}$

Since$9>6$and $9$is in the $y^2$term, the major axis will be vertical.

The value is $a^2=6$.

This implies that $a=Â\pm \sqrt{6}$.

This follows that the minor axis is horizontal and the distance from the center to the vertices is $\sqrt{6}$.

The value is $b^2=9$.

This implies that $b=Â\pm \sqrt{9}$.

This follows that the distance from the center to the endpoints of the major axis is $\sqrt{9}$.

The graph of an ellipse is shown below: