91影视

The given ellipse is $9x^2+16y^2=144$.

The standard form of an equation is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1鈥�\left(1\right)$.

From $\left(1\right)$,observe thatthe equation $9x^2+16y^2=144$is not in the standard form since both the $x$and $y$terms are squared and have different coefficients.

So, divide the equationwidth="119">9x2+16y2=144both sides by $144$to get the equation in standard form.

$\begin{array}{rcl}\frac{9x^2}{144}+\frac{16y^2}{144}& =& \frac{144}{144}\\ \frac{x^2}{16}+\frac{y^2}{9}& =& 1\end{array}$

The equation$\frac{x^2}{16}+\frac{y^2}{9}=1$is in standard form centered at the origin.

So, compare the equation $\left(1\right)$with $\frac{x^2}{16}+\frac{y^2}{9}=1$to find $a^2$and $b^2$.

$\begin{array}{rcl}{a}^2& =& 16\\ b^2& =& 9\end{array}$

Since $16>9$and $16$is in the $x^2$term, the major axis will be horizontal.

The endpoints will be the $x$-intercepts.

Since$a^2=16$,$a=卤4$.

Thus, the coordinates of the endpoints of the major axis are $(4,0)$and $(-4,0)$.

The endpoints will be the $y$-intercepts.

Since $b^2=9$, $b=卤3$.

Thus, the coordinates of the endpoints of the minor axis are $(0,3)$and $(0,-3)$.

The graph of an ellipse is shown below: