Graph the ellipse ( x − 1 ) 2 9 + ( y + 3 ) 2 4 = 1 \frac{\left(x-1\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1 9 ( x − 1 ) 2 + 4 ( y + 3 ) 2 = 1 .

Hey, everyone. In this problem, we need to graph the following ellipse based on the equation. So to solve this problem, what we can do is first determine what the orientation of the ellipse is going to be. And since I can see that our larger number in the denominator 9 is underneath the x value, that means we're going to have a horizontal ellipse of some kind. And the equation for a horizontal ellipse is x minus h squared over a squared plus y minus k squared over b squared is all equal to 1. So this is the general form, and we have a squared underneath x since we know it's a horizontal ellipse. Now what we need to do from here is figure out where the center of our ellipse is going to be. Well I can see that this h corresponds with positive 1. So that means horizontally we're going to be shifted over 1 unit. I can see that k corresponds with this 3, but notice that there is a minus or excuse me, a plus sign instead of a minus sign. And because of that, that means k is actually going to be negative 3. So we need to be shifted down 1, 2, 3 units. So the center of our ellipse is going to be right about here at 1 negative 3. Now, from here what we need to do is figure out what the size of this ellipse is going to be. And to figure this out, we can look at our equation and see these numbers in the denominator and know how they are associated with the major and minor axes. Now recall that the larger number is the major axis squared, and the smaller number is the minor axis squared. So for the major axis, we can see that a squared is equal to 9, meaning that a is going to be the square root of 9 which is 3. So So what I can do is go over here to our ellipse. And since I see that we're dealing with a horizontal ellipse, that means we need to go horizontally to the left and to the right 3 units. So I can go 1, 2, 3 units to the left, and 1, 2, 3 units to the right. Now I can also see that our minor axis is or our minor axis squared is 4. So if b squared is equal to 4, and that means b is going to be the square root of 4, which is 2. So starting from the center point I can go up 2 units and I can go down 2 units and this will give me the 2 points for our minor axis. Now at this point, all I need to do is connect these outside points with a smooth curve, which looks something like that. So once we connect these points, this is going to give us the ellipse that we're looking for. And that right there is the graph of the ellipse and the answer to the problem. So thanks for watching and let's move on.

19. Conic Sections

Ellipses: Standard Form

Precalculus

19. Conic Sections

Ellipses: Standard Form

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Multiple Choice

Graph the ellipse $\frac{\left(x-1\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$ .

option a

option b

option c

optiond

Verified step by step guidance

Identify the standard form of the ellipse equation: $ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 $, where $(h, k)$ is the center of the ellipse.

From the given equation $ \frac{(x-1)^2}{9} + \frac{(y+3)^2}{4} = 1 $, identify the center of the ellipse as $(h, k) = (1, -3)$.

Determine the values of $a^2$ and $b^2$ from the denominators: $a^2 = 9$ and $b^2 = 4$. Thus, $a = 3$ and $b = 2$.

Since $a > b$, the major axis is horizontal. The length of the major axis is $2a = 6$ and the length of the minor axis is $2b = 4$.

Plot the center of the ellipse at $(1, -3)$, then draw the ellipse extending 3 units horizontally from the center and 2 units vertically from the center.

5:12m

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Graph the ellipse $\frac{\left(x-1\right)^2}{9}+\frac{\left(y+3\right)^2}{4}=1$ .