What are the foci of the ellipse? Graph the ellipse. 18x^2 + 36y^2 =648
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).
To find the foci of the ellipse, we first need to rewrite the equation in standard form, which is (x-h)^2/a^2 + (y-k)^2/b^2 = 1 for horizontal ellipses or (x-h)^2/b^2 + (y-k)^2/a^2 = 1 for vertical ellipses.
For the given equation, 18x^2 + 36y^2 = 648, we can start by dividing both sides of the equation by 648 to get the equation in standard form:
18x^2/648 + 36y^2/648 = 1
Simplifying further:
x^2/36 + y^2/18 = 1
Now, we can see that this is an ellipse with a^2 = 36 and b^2 = 18. To find the value of a and b, we take the square root:
a = sqrt(36) = 6
b = sqrt(18) = 3√2
The foci of an ellipse are given by the equation c^2 = a^2 - b^2, where c is the distance between the center and the foci.
So, c^2 = 6^2 - (3√2)^2
c^2 = 36 - 18
c^2 = 18
c = √18 = 3√2
To graph the ellipse, we can start by locating the center, which is at the point (h, k) = (0, 0). From the center, we can plot the major and minor axes. The major axis has a length of 2a = 2(6) = 12 and is along the x-axis, while the minor axis has a length of 2b = 2(3√2) and is along the y-axis.
Then, we can plot the foci. Since the center is at (0, 0), the foci will be at (±c, 0), which gives us the points (±3√2, 0).
Now we are ready to graph the ellipse. Using the information above, we can plot the center, the foci, and draw the ellipse using the major and minor axes as guides.
To find the foci of the ellipse, we need to determine the major and minor axes first.
Step 1: Divide the equation by 648 to simplify it:
(x^2)/36 + (y^2)/18 = 1
Step 2: Compare the equation with the standard form of an ellipse:
(x^2)/(a^2) + (y^2)/(b^2) = 1
From the standard form, we can see that a^2 = 36 and b^2 = 18.
Step 3: Take the square root of both a^2 and b^2 to find the values of a and b:
a = √36 = 6
b = √18 ≈ 4.24 (rounded to 2 decimal places)
The major axis (2a) of the ellipse is 2 * 6 = 12 units, while the minor axis (2b) is 2 * 4.24 ≈ 8.48 units.
Step 4: Determine the location of the center of the ellipse, which is the origin (0,0).
Step 5: Find the distance from the center of the ellipse to the foci using the formula:
c = √(a^2 - b^2)
c = √(36 - 18) = √18 ≈ 4.24 (rounded to 2 decimal places)
There are two foci located on the major axis of the ellipse, equidistant from the origin. The foci are on the y-axis, so their coordinates are (0, ±c) = (0, ±4.24) approximated to two decimal places.
Step 6: Graphing the ellipse:
- Plot the center (0,0) on the coordinate plane.
- Draw the major axis with length 2a = 12 units, extending 6 units in both directions from the center along the x-axis.
- Draw the minor axis with length 2b = 8.48 units, extending 4.24 units in both directions from the center along the y-axis.
- Plot the two foci on the y-axis at (0, ±4.24).
The graph should resemble a vertically stretched ellipse centered at the origin with foci at (0, 4.24) and (0, -4.24).