Determine the vertices and foci of the following ellipse: x249+y236=1\frac{x^2}{49}+\frac{y^2}{36}=149x2+36y2=1.

Vertices: (7,0),(−7,0)\left(7,0\right),\left(-7,0\right)(7,0),(−7,0)Foci: (13,0),(−13,0)\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)(13,0),(−13,0)

Determine the vertices and foci of the following ellipse: x 2 49 + y 2 36 = 1 \frac{x^2}{49}+\frac{y^2}{36}=1 49 x 2 + 36 y 2 = 1 .

Hey everyone. Let's give this problem a try. So in this problem, we are asked to determine the vertices and foci of the following ellipse, and we're given this equation. Now first off, by looking at this equation, I noticed that the larger number is underneath the x squared, And the larger number is associated with the semi major axis. This is a squared. Whereas the smaller number is going to be b squared. So because I see that a squared is underneath the x squared, I know that we are dealing with a horizontal ellipse. Now in order to find the vertices, the vertices are just going to be the distance of a. So since I see that a squared is equal to 49, I can find the a value by taking the square root on both sides of this little equation here. This is gonna be that a is the square root of 49, which is seven. So So what that means is that our vertices are going to be at seven zero and at negative seven zero. Because if you think about it, if we have this kind of ellipse shape and you can imagine that we have some graph of an ellipse, this is a horizontally oriented ellipse. And the vertices are going to be at these two endpoints. So one would be at negative seven zero, and the other one would be at positive seven zero. So those are the vertices. Now we also need to find the foci. And recall we can find the foci by traveling a distance c. But to find c, we need to use the equation for c, which is c squared is equal to a squared minus b squared. Now for a squared, we determined that a squared was what's under the x squared which is 49, and b squared is what's under the y squared which is 36. So we're going to have that c squared is 49 minus 36, and we can do that math over here. Six nine minus six is three, and then four minus three is one. So 49 minus 36 is 13, and then if I take the square root on both sides of this little equation, we can cancel those giving us that c is equal to the square root of 13. And this doesn't simplify down any farther, so this would be our answer right there. Meaning that our foci are going to be at the square root of 13, and then at the negative square root of 13. So based on what we have for this equation these would be the vertices and these would be the foci. Hope you found this video helpful, thanks for watching.

Determine the vertices and foci of the following ellipse: x249+y236=1\frac{$$x^2$$}{49}+\frac{$$y^2$$}{36}=149x2+36y2=1.

Vertices: (7,0),(−7,0)\left(7,0\right),\left(-7,0\right)(7,0),(−7,0)Foci: (13,0),(−13,0)\left(\sqrt{13},0\right),\left(-\sqrt{13},0\right)(13,0),(−13,0)

Vertices: (7,0),(−7,0)\left(7,0\right),\left(-7,0\right)(7,0),(−7,0)Foci: (6,0),(−6,0)\left(6,0\right),\left(-6,0\right)(6,0),(−6,0)

Vertices: (6,0),(−6,0)\left(6,0\right),\left(-6,0\right)(6,0),(−6,0)Foci: (7,0),(−7,0)\left(7,0\right),\left(-7,0\right)(7,0),(−7,0)

Vertices: (0,7),(0,−7)\left(0,7\right),\left(0,-7\right)(0,7),(0,−7)Foci: (0,13),(0,−13)\left(0,\sqrt{13}\right),\left(0,-\sqrt{13}\right)(0,13),(0,−13)

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

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Multiple Choice

Determine the vertices and foci of the following ellipse: $$\frac{x^2}{49}$+$\frac{y^2}{36}$=1$ .

Vertices: $$\left$(7,0$\right$),$\left$(-7,0$\right$)$

Foci: $$\left$(6,0$\right$),$\left$(-6,0$\right$)$

Vertices: $$\left$(6,0$\right$),$\left$(-6,0$\right$)$

Foci: $$\left$(7,0$\right$),$\left$(-7,0$\right$)$

Vertices: $$\left$(7,0$\right$),$\left$(-7,0$\right$)$

Foci: $$\left$($\sqrt{13}$,0$\right$),$\left$(-$\sqrt{13}$,0$\right$)$

Vertices: $$\left$(0,7$\right$),$\left$(0,-7$\right$)$

Foci: $$\left$(0,$\sqrt{13}\[\right$),$\left$(0,-$\sqrt{13}\]\right$)$

Verified step by step guidance

Step 1: Recognize the equation of the ellipse: $ \frac{x^2}{49} + \frac{y^2}{36} = 1 $. This is in the standard form $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a^2 $ and $ b^2 $ represent the squares of the semi-major and semi-minor axes respectively.

Step 2: Identify $ a^2 = 49 $ and $ b^2 = 36 $. Calculate $ a $ and $ b $ by taking the square root: $ a = \sqrt{49} = 7 $ and $ b = \sqrt{36} = 6 $. Since $ a > b $, the ellipse is horizontal, meaning the major axis is along the x-axis.

Step 3: Determine the vertices. For a horizontal ellipse, the vertices are located at $ (\pm a, 0) $. Substituting $ a = 7 $, the vertices are $ (7, 0) $ and $ (-7, 0) $.

Step 4: Calculate the distance to the foci using the formula $ c = \sqrt{a^2 - b^2} $. Substituting $ a^2 = 49 $ and $ b^2 = 36 $, $ c = \sqrt{49 - 36} = \sqrt{13} $. The foci are located at $ (\pm c, 0) $, which are $ (\sqrt{13}, 0) $ and $ (-\sqrt{13}, 0) $.

Step 5: Summarize the results. The vertices of the ellipse are $ (7, 0) $ and $ (-7, 0) $. The foci are $ (\sqrt{13}, 0) $ and $ (-\sqrt{13}, 0) $.

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Determine the vertices and foci of the following ellipse: x249+y236=1\(\frac{x^2}{49}\)+\(\frac{y^2}{36}\)=149x2+36y2=1.

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Determine the vertices and foci of the following ellipse: $\(\frac{x^2}{49}\)+\(\frac{y^2}{36}\)=1$ .