Determine the vertices and foci of the ellipse (x+1)2+(y−2)24=1\left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1(x+1)2+4(y−2)2=1.

Vertices: ( − 1 , 4 ) , ( − 1 , 0 ) \left(-1,4\right),\left(-1,0\right) ( − 1 , 4 ) , ( − 1 , 0 ) Foci: ( − 1 , 2 + 3 ) , ( − 1 , 2 − 3 ) \left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right) ( − 1 , 2 + 3 <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"> ) , ( − 1 , 2 − 3 <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 ellipse ( x + 1 ) 2 + ( y − 2 ) 2 4 = 1 \left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 1 .

Welcome back, everyone. So let's see if we can solve this problem. And I will warn you that this problem is actually a bit tedious. So here we are asked to determine the vertices and foci for the ellipse, and we are given the equation. Now to solve this problem, I think it's really best to try and understand what the values are going to be for the vertices and foci based on how the graph would look and how the equation relates to the graph. So what I'm going to do is write a mini graph over here for reference, and we're not gonna put a lot of detail into this just to kind of show you what the ellipse is gonna look like, And I'm also going to rewrite this equation. So we're going to have x plus one one squared over 1 plus, and then we're gonna have y minus 2 squared over 4 equals 1. This is all just to make things a bit more clear as we solve this problem. Now what I'm first gonna do is determine the center of the ellipse. And if I look at the numbers we have here we have a positive one, or we have plus 1, and then we have minus 2. And recall that the general form has x plus or x minus h squared over, and then in this case since we see that this is the smaller number this would be b squared, and then we'd have plus y minus k squared, and then this is the larger number so that's gonna be over a squared, and that whole thing equals 1. Now because we have a plus one that means h is gonna be negative one since we have a plus sign rather than a minus sign. And since we have a minus 2, that means that k is going to be positive 2. So this is gonna be our h and k, meaning the center of our ellipse would be at negative 12. So this would be the center. Now at this point, what we need to do and keep in mind, our ultimate goal is to determine the vertices and foci. So since we've now found the center of our ellipse, we know that this is going to be a vertically oriented ellipse since we have the 4, the larger number underneath the y as we discussed before. So the ellipse is gonna be something like this. Now I can see here that a squared is associated with this 4. So that means a squared is 4 and then we take the square root of both sides that gives us a which is 2. So if I go up 1, 2 units, and then down 1, 2 units down here, that's going to give me the 2 major axes, which are also going to correspond with the 2 vertices of the ellipse. And then if I use my b value, b squared is equal to 1, and if b squared is 1, then b is gonna be the square root of 1, which is just 1. So if I go to the right, 1 unit, and then to the left, 1 unit, the ellipse is gonna end up looking something like that. Now I'm ultimately trying to find the vertices, and the vertices are right here and right there. And notice just graphically, one of the vertex points is going to end up at this point, which is at negative 1, and vertically it's up at 1, 2, 3, 4. So one of our points is going to be at negative 14, and the other vertex point is going to be down here at negative 10, because we're at the 0 point on the y axis. So these are going to be the 2 vertices of our ellipse. So we have one point at negative one four, and we have another point at negative one zero. Now something else you could have done to solve this is recognize that the vertices are going to be well, since we have a vertical ellipse, then the horizontal piece h is gonna stay constant. But the vertical piece, we need to add and subtract a from, and that would give us the 2 vertex points. Because since we're going up and down, it's the a value we add and subtract to whatever the vertical position is of the center of the ellipse. But that's just something if you like just kinda remembering coordinates. I personally like to remember this with the graph, but, you know, it really just depends on your preference there. So we've now found the 2 the 2 vertices for our ellipse. We also need to find the foci. Well, following the trend of where the foci end up on the ellipse, they're probably gonna end up being somewhere here and there, just generally in that region, and I'm actually gonna do this in green since we already used blue. So this is gonna be where the foci end up, and recall that for the foci that's going to be c squared is equal to a squared minus b squared, where c is the distance from the center to the 2 foci. And I can see here that a squared well we determined that a squared is associated with this 4. So that means c squared is equal to 4 minus b squared which is 1. So we have 4 minuteus 1 that means that c squared is going to be 3, giving us that c is equal to the square root of 3 when we take the square root on both sides. So this is our c value. Now if we start at the center at this point, this is again a vertical ellipse that we're dealing with. So that means for the foci, and I'm gonna do this in green again. So the foci are going to be the same horizontal position. We don't move horizontally, but vertically we're going to add and subtract c to get each of our distances. So to find the 2 foci of this ellipse, well, we're going we have the same h value, so h is at negative one, and then our focus point is going to be our k value, which we said was 2, and this is going to be plus the c value, which is the square root of 3. Then we're going to have the other focus point at negative one, and then we're going to have 2 minus the square root of 3. And this right here would be the foci of our ellipse. So notice that the foci just go up and down, because we figured out that this is a vertical ellipse, and that the vertices also go up and down based on the ellipses orientation. This is how you can find the vertices and foci of an ellipse. So hope you found this helpful. Thanks for watching, and let's move on.

Determine the vertices and foci of the ellipse ( x + 1 ) 2 + ( y − 2 ) 2 4 = 1 \left(x+1\right)^2+\frac{\left(y-2\right)^2}{4}=1 ( x + 1 ) 2 + 4 ( y − 2 ) 2 = 1 .

Vertices: ( − 1 , 4 ) , ( − 1 , 0 ) \left(-1,4\right),\left(-1,0\right) ( − 1 , 4 ) , ( − 1 , 0 ) Foci: ( − 1 , 2 + 3 ) , ( − 1 , 2 − 3 ) \left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right) ( − 1 , 2 + 3 <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"> ) , ( − 1 , 2 − 3 <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: ( − 1 , 4 ) , ( − 1 , 0 ) \left(-1,4\right),\left(-1,0\right) ( − 1 , 4 ) , ( − 1 , 0 ) Foci: ( − 2 , 2 ) , ( 0 , 2 ) \left(-2,2\right),\left(0,2\right) ( − 2 , 2 ) , ( 0 , 2 )

Vertices: ( − 2 , 2 ) , ( 0 , 2 ) \left(-2,2\right),\left(0,2\right) ( − 2 , 2 ) , ( 0 , 2 ) Foci: ( 1 , 2 + 3 ) , ( 1 , 2 − 3 ) \left(1,2+\sqrt3\right),\left(1,2-\sqrt3\right) ( 1 , 2 + 3 <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"> ) , ( 1 , 2 − 3 <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: ( − 2 , 2 ) , ( 0 , 2 ) \left(-2,2\right),\left(0,2\right) ( − 2 , 2 ) , ( 0 , 2 ) Foci: ( 2 + 3 , 1 ) , ( 2 − 3 , 1 ) \left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right) ( 2 + 3 <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"> , 1 ) , ( 2 − 3 <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"> , 1 )

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

Vertices: $\left(-1,4\right),\left(-1,0\right)$

Foci: $\left(-1,2+\sqrt3\right),\left(-1,2-\sqrt3\right)$

Vertices: $\left(-1,4\right),\left(-1,0\right)$

Foci: $\left(-2,2\right),\left(0,2\right)$

Vertices: $\left(-2,2\right),\left(0,2\right)$

Foci: $\left(1,2+\sqrt3\right),\left(1,2-\sqrt3\right)$

Vertices: $\left(-2,2\right),\left(0,2\right)$

Foci: $\left(2+\sqrt3,1\right),\left(2-\sqrt3,1\right)$

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, $a$ is the semi-major axis, and $b$ is the semi-minor axis.

Rewrite the given equation $(x+1)^2 + \frac{(y-2)^2}{4} = 1$ in standard form. Here, $h = -1$, $k = 2$, $a = 2$, and $b = 1$.

Determine the vertices by using the formula $(h, k \pm a)$ for a vertical ellipse. Substitute $h = -1$, $k = 2$, and $a = 2$ to find the vertices.

Calculate the foci using the formula $(h, k \pm c)$, where $c = \sqrt{a^2 - b^2}$. Compute $c$ using $a = 2$ and $b = 1$.

Substitute $h = -1$, $k = 2$, and $c$ into the foci formula to find the coordinates of the foci.

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