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 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 0 ) , ( − 13 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 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 7. So what that means is that our vertices are going to be at 70 and at negative 70. 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. 6 9 minuteus 6 is 3, and then 4 minuteus 3 is 1, so 49 minuteus 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 further so this would be our answer right there, meaning that our foci are going to be at the square root of 130, and then at the negative square root of 130. 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: x 2 49 + y 2 36 = 1 \frac{x^2}{49}+\frac{y^2}{36}=1 49 x 2 + 36 y 2 = 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 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 0 ) , ( − 13 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> , 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 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> ) , ( 0 , − 13 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

8. Conic Sections

Ellipses: Standard Form

College Algebra

8. Conic Sections

Ellipses: Standard Form

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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

Identify the standard form of the ellipse equation: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. In this problem, the equation is $ \frac{x^2}{49} + \frac{y^2}{36} = 1 $.

Determine the values of $ a^2 $ and $ b^2 $ from the equation. Here, $ a^2 = 49 $ and $ b^2 = 36 $.

Calculate $ a $ and $ b $ by taking the square roots: $ a = \sqrt{49} = 7 $ and $ b = \sqrt{36} = 6 $.

Since $ a > b $, the major axis is along the x-axis. The vertices are at $ (\pm a, 0) $, which are $ (7, 0) $ and $ (-7, 0) $.

To find the foci, use the formula $ c^2 = a^2 - b^2 $. Calculate $ c = \sqrt{49 - 36} = \sqrt{13} $. The foci are at $ (\pm c, 0) $, which are $ (\sqrt{13}, 0) $ and $ (-\sqrt{13}, 0) $.

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