What are the foci of the ellipse? Graph the ellipse. 18x^2 + 36y^2 =648

Answer 1

To find the foci of the ellipse, we first need to rewrite the given equation in the standard form of an ellipse, which is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

where (a,b) are the semi-major and semi-minor axes.

Divide both sides of the equation by 648, to isolate the terms and rewrite it in standard form:

$\frac{x^2}{36} + \frac{y^2}{18} = 1$

Now we can see that $a^2 = 36$ and $b^2 = 18$, so a = 6 and b = $\sqrt{18}=3\sqrt{2}$.

To find the foci, we use the relationship $c^2 = a^2 - b^2$, where c is the distance from the center of the ellipse to each focus.

We find that:

$c^2 = 36 - 18 = 18$

$c = \sqrt{18}= 3\sqrt{2}$

Since the larger denominator is under the $x^2$ term, it means that the ellipse is stretched along the x-axis, and thus its foci will have a form (±c,0). Therefore, the foci of the ellipse are ($\pm3\sqrt{2}$, 0).

To graph the ellipse, plot the center at (0,0) and draw an ellipse with semi-major axis of length 6 along the x-axis and semi-minor axis of length $3\sqrt{2}$ along the y-axis. Don't forget to mark the foci at ($\pm3\sqrt{2}$, 0).