Determine the vertices and foci of the following ellipse: x29+y216=1\frac{x^2}{9}+\frac{y^2}{16}=19x2+16y2=1.

Vertices: ( 0 , 4 ) , ( 0 , − 4 ) \left(0,4\right),\left(0,-4\right) ( 0 , 4 ) , ( 0 , − 4 ) Foci: ( 0 , 7 ) , ( 0 , − 7 ) \left(0,\sqrt7\right),\left(0,-\sqrt7\right) ( 0 , 7 <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 , − 7 <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"> )

Determine the vertices and foci of the following ellipse: x 2 9 + y 2 16 = 1 \frac{x^2}{9}+\frac{y^2}{16}=1 9 x 2 + 16 y 2 = 1 .

Welcome back everyone. So let's see if we can solve this problem. We are asked to determine the vertices and foci of the following ellipse, x squared over 9 plus y squared over 16 equals 1. Now to solve this problem, our first step should be to figure out how this ellipse is going to be oriented. Are we going to have a horizontal ellipse, or are we going to have a vertical ellipse? Well, if we look at this equation, I need to find where the larger number is in the denominator, and I see that the larger number is 16. So since the larger number is underneath the y squared, that means we're going to have a vertical ellipse. Because if we're underneath the y squared, that means that our ellipse is going to be stretched on the y axis, so it's going to look something like this. So that means that 16 this number that we see that is associated with the semi major axis. So 16 is going to be a squared, whereas the smaller number is going to be the semi minor axis or b square. So now that we have these, we can go ahead and find the vertices first. I can find the vertices by solving for a because a will tell us the distance to either vertex point. Now you see that a squared is equal to 16 and if I take the square root on both sides of this equation we'll get that a is the square root of 16 which is 4. So that means that our vertices are going to be at 40 or excuse me, 4 and negative 4. So writing up the vertices, we're going to have 0 4, and 0, negative 4. And notice that we're making the fours the second coordinate because we're on the y axis this time. So those are the vertices. But now we need to find the foci. And to find the foci, we need c. And to calculate c, we need to recognize that c squared is a squared minus b squared. Well, we can see that a squared is going to be this 16, and we can see that b squared is going to be this 9. So we'll have 16 minuteus 9 which is 7, and if I take the square root on both sides of this equation we'll get that c is equal to the square root of 7, and this is the distance to the 2 foci which will be about there and there let's just say. So that means that our foci are going to be at 0 square root of 7 and at 0 negative square root of 7. So this is how we can find the vertices and the foci for our lips. So that's the answer to this problem, hope you found this video helpful.

Determine the vertices and foci of the following ellipse: x 2 9 + y 2 16 = 1 \frac{x^2}{9}+\frac{y^2}{16}=1 9 x 2 + 16 y 2 = 1 .

Vertices: ( 0 , 4 ) , ( 0 , − 4 ) \left(0,4\right),\left(0,-4\right) ( 0 , 4 ) , ( 0 , − 4 ) Foci: ( 0 , 7 ) , ( 0 , − 7 ) \left(0,\sqrt7\right),\left(0,-\sqrt7\right) ( 0 , 7 <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 , − 7 <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"> )

Vertices: ( 4 , 0 ) , ( − 4 , 0 ) \left(4,0\right),\left(-4,0\right) ( 4 , 0 ) , ( − 4 , 0 ) Foci: ( 7 , 0 ) , ( − 7 , 0 ) \left(\sqrt7,0\right),\left(-\sqrt7,0\right) ( 7 <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 ) , ( − 7 <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: ( 4 , 0 ) , ( − 4 , 0 ) \left(4,0\right),\left(-4,0\right) ( 4 , 0 ) , ( − 4 , 0 ) Foci: ( 3 , 0 ) , ( − 3 , 0 ) \left(3,0\right),\left(-3,0\right) ( 3 , 0 ) , ( − 3 , 0 )

Vertices: ( 0 , 4 ) , ( 0 , − 4 ) \left(0,4\right),\left(0,-4\right) ( 0 , 4 ) , ( 0 , − 4 ) Foci: ( 0 , 3 ) , ( 0 , − 3 ) \left(0,3\right),\left(0,-3\right) ( 0 , 3 ) , ( 0 , − 3 )

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}{9}$+$\frac{y^2}{16}$=1$ .

Vertices: $$\left$(4,0$\right$),$\left$(-4,0$\right$)$

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

Vertices: $$\left$(0,4$\right$),$\left$(0,-4$\right$)$

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

Vertices: $$\left$(4,0$\right$),$\left$(-4,0$\right$)$

Foci: $$\left$(3,0$\right$),$\left$(-3,0$\right$)$

Vertices: $$\left$(0,4$\right$),$\left$(0,-4$\right$)$

Foci: $$\left$(0,3$\right$),$\left$(0,-3$\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}{9} + \frac{y^2}{16} = 1 $.

Determine the values of $ a^2 $ and $ b^2 $ from the equation. Here, $ a^2 = 9 $ and $ b^2 = 16 $.

Since $ b^2 > a^2 $, the major axis is vertical. Calculate $ a $ and $ b $ by taking the square roots: $ a = 3 $ and $ b = 4 $.

Find the vertices of the ellipse. For a vertical major axis, the vertices are at $ (0, \pm b) $. Thus, the vertices are $ (0, 4) $ and $ (0, -4) $.

Calculate the foci using the formula $ c^2 = b^2 - a^2 $. Here, $ c^2 = 16 - 9 = 7 $, so $ c = \sqrt{7} $. The foci are at $ (0, \pm \sqrt{7}) $.

5:12m

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Struggling with College Algebra?

Determine the vertices and foci of the following ellipse: x29+y216=1\(\frac{x^2}{9}\)+\(\frac{y^2}{16}\)=19x2+16y2=1.

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Ellipses: Standard Form practice set

Determine the vertices and foci of the following ellipse: $\(\frac{x^2}{9}\)+\(\frac{y^2}{16}\)=1$ .