## anonymous one year ago (x-7)^2/4 + (y-3)^2/9=1 How would I graph that?

1. misty1212

HI!!

2. misty1212

ellipse right?

3. misty1212

looks like this http://www.wolframalpha.com/input/?i=+%28x-7%29%5E2%2F4+%2B+%28y-3%29%5E2%2F9%3D1 you need steps?

4. anonymous

So, I assume that you know your equation is an ellipse. The general equation of an ellipse is: $\frac{ (x-h)^{2} }{ a^{2} } + \frac{(y-k)^{2}}{b^{2}} = 1$ In general, the $$a^{2}$$ is the largest of the 2 denominators and the $$b^{2}$$ is the smallest of the two denominators, so $$a^{2}$$ and $$b^{2}$$ may be flip-flopped in that general equation. That doesn't matter for this problem, though, just a notation explanation. The center of an ellipse is at the coordinate (h,k). If you take the denominator of the fraction with x's and square root it, you can find the distance from the center to the vertices left and right from the center (as in along the x-axis). So here's what I mean by that: In your equation, you have a center at (7,3). The denominator of the fraction containing x's is 4. If I square root that, Ill get 2. This number 2 is how far I need to travel left and right from the center to find two points on the ellipse. Therefore one point on the ellipse is at (5,3) and the other is at (9,3). If we take the square root of the denominator of the fraction containing x's, we can find the distance from the center to the vertices up and down from the center (as in along the y-axis). Since the denominator of that fraction is 9, the square root of that is 3 and thus the distance from the center to two more points is 3. So going up and down 3 units from the center, we find the points (7,0) and (7,6). With those 4 points, we can graph the ellipse |dw:1433357061917:dw| If needed, you can also find the coordinates of the foci of the ellipse. For an ellipse, the foci are located a distance of c from the center oriented in the directions of the bigger denominator. As in if the x terms have a bigger denominator, then the foci will be left and right from the center. If the y terms have a bigger denominator, the foci will be up and down from the center. Or visually, you can think of the foci as being in the directions where the ellipse stretches more. This distance c is found from this relation : $a^2 - b^2 = c^2$ Again, $$a^{2}$$ is the bigger of the two denominators and $$b^{2}$$ is the smallest. So for us, $$a^{2} = 9$$ and $$b^{2} = 4$$. Thus we have: $$a^{2} - b^{2} = c^{2}$$ $$9-4 = c^{2}$$ $$5 = c^{2}$$ $$c = \sqrt{5}$$ SInce the bigger denominator is with the y's, we find the foci by moving up and down $$\sqrt{5}$$ units from the center. Thus we have foci at the points $$(7,3+\sqrt{5})$$ and $$(7, 3-\sqrt{5})$$. It's a lot of info, but hopefully it makes sense and it helps.

5. anonymous

Thank you!