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),(−1,2−3)

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 kinda 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 one plus, and then we're gonna have y minus two squared over four equals one. 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 o positive one or we have plus one and then we have minus two. 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 one. 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 two, that means that k is going to be positive two. So this is gonna be our h and k, meaning the center of our ellipse would be at negative one two. 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 four, 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 four. So that means a squared is four and then we take the square root of both sides that gives us a which is two. So if I go up one, two units and then down one, two units down here, that's going to give me the two major axes, which are also going to correspond with the two vertices of the ellipse. And then if I use my b value, b squared is equal to one. And if b squared is one, then b is gonna be the square root of one, which is just one. So if I go to the right one unit and then to the left one 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 one, and vertically it's up at one, two, three, four. So one of our points is going to be at negative one, four. And the other vertex point is going to be down here at negative one, zero, because we're at the zero point on the y axis. So these are gonna be the two 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 two 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 two the two 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 two foci. And I can see here that a squared well, we determined that a squared is associated with this four. So that means c squared is equal to four minus b squared, which is one. So we have four minus one. That means that c squared is gonna be three, giving us that c is equal to the square root of three 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 two foci of this ellipse, well, we're going we have the same h value, so h is at negative one. And then our foci is going to be our k value, which we said was two. And this is going to be plus the c value, which is the square root of three. But then we're going to have the other focus point at negative one, and then we're going to have two minus the square root of three. 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)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),(−1,2−3)

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),(1,2−3)

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+3,1),(2−3,1)

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 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$(-1,2+$\sqrt$3$\right$),$\left$(-1,2-$\sqrt$3$\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$(1,2+$\sqrt$3$\right$),$\left$(1,2-$\sqrt$3$\right$)$

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

Foci: $$\left$(2+$\sqrt$3,1$\right$),$\left$(2-$\sqrt$3,1$\right$)$

Verified step by step guidance

Step 1: Recognize the equation of the ellipse. The given equation $(x+1)^2 + \frac{(y-2)^2}{4} = 1$ is in the standard form of an ellipse centered at $(-h, k)$, where $h = -1$ and $k = 2$.

Step 2: Identify the orientation of the ellipse. Since the coefficient of $(x+1)^2$ is 1 and the coefficient of $(y-2)^2$ is $\frac{1}{4}$, the major axis is vertical, and the minor axis is horizontal.

Step 3: Determine the lengths of the semi-major axis $a$ and semi-minor axis $b$. From the equation, $a^2 = 4$ (denominator under $y$) and $b^2 = 1$ (denominator under $x$). Thus, $a = 2$ and $b = 1$.

Step 4: Calculate the vertices. The vertices lie along the major axis, which is vertical. Using the center $(-1, 2)$ and $a = 2$, the vertices are $(-1, 2 + 2)$ and $(-1, 2 - 2)$, or $(-1, 4)$ and $(-1, 0)$.

Step 5: Determine the foci. The foci are located along the major axis at a distance $c$ from the center, where $c = \sqrt{a^2 - b^2}$. Substituting $a^2 = 4$ and $b^2 = 1$, $c = \sqrt{4 - 1} = \sqrt{3}$. Thus, the foci are $(-1, 2 + \sqrt{3})$ and $(-1, 2 - \sqrt{3})$.

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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.

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