91Ó°ÊÓ

Consider the equation.

$4x^2+y^2−16x−6y+9=0$

Group the $x$terms and $y$terms.

$(4x^2−16x)+(y^2−6y)=-9$

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

$4(x^2−4x)+(y^2−6y)=-9$

Complete the squares.

$\begin{array}{rcl}4(x^2−4x\mathbf{+}\mathbf{4})+(y^2−6y\mathbf{+}\mathbf{9})& =& -9\mathbf{+}\mathbf{16}\mathbf{+}\mathbf{9}\\ 4{(x-2)}^2+{(y-3)}^2& =& 16\end{array}$

$\begin{array}{rcl}\frac{4{(x-2)}^2}{16}+\frac{{(y-3)}^2}{16}& =& \frac{16}{16}\\ \frac{{(x-2)}^2}{4}+\frac{{(y-3)}^2}{16}& =& 1\end{array}$

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

Compare the given equation with standard form and conclude that$\begin{array}{rcl}\frac{{(x-2)}^2}{4}+\frac{{(y-3)}^2}{16}& =& 1\end{array}$is in standard form and the center is $(2,3)$.

Compare the equation $\left(1\right)$with $\begin{array}{rcl}\frac{{(x-2)}^2}{4}+\frac{{(y-3)}^2}{16}& =& 1\end{array}$to find $a^2$and $b^2$.

$\begin{array}{rcl}{a}^2& =& 4\\ b^2& =& 16\end{array}$

Since $16>4$and $16$is in the $y^2$term, the major axis will be vertical.

The value is$a^2=4$.

This implies that $a=Â\pm 2$.

This follows that the minor axis is horizontal and the distance from the center to the vertices is width="10">2.

The value is $b^2=16$.

This implies that $b=Â\pm 4$.

This follows that the distance from the center to the endpoints of the major axis is $4$.

The graph of an ellipse is shown below: