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),(0,−7)

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 nine plus y squared over 16 equals one. 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 are 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 squared. 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 can 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 four. So that means that our vertices are going to be at four zero or excuse me zero four and negative zero four. So writing up the vertices, we're going to have zero four and zero negative four. 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 nine. So we'll have 16 minus nine, which is seven. 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 seven. And this is the distance to the two foci which will be about there and there let's just say. So that means that our foci are going to be at zero square root of seven and at zero negative square root of seven. 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: 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),(0,−7)

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)(7,0),(−7,0)

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)

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}{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

Step 1: Recognize the equation of the ellipse in standard form: $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $. Here, $ a^2 = 16 $ and $ b^2 = 9 $. Identify $ a $ and $ b $ as the square roots of these values: $ a = 4 $ and $ b = 3 $.

Step 2: Determine the orientation of the ellipse. Since $ a^2 > b^2 $, the major axis is vertical, and the ellipse is taller than it is wide. The center of the ellipse is at $ (0, 0) $.

Step 3: Calculate the vertices. The vertices lie along the major axis, which is vertical. The coordinates of the vertices are $ (0, a) $ and $ (0, -a) $, resulting in $ (0, 4) $ and $ (0, -4) $.

Step 4: Compute the foci. The foci are located along the major axis at a distance $ c $ from the center, where $ c = \sqrt{a^2 - b^2} $. Substitute $ a^2 = 16 $ and $ b^2 = 9 $ into $ c = \sqrt{a^2 - b^2} $ to find $ c = \sqrt{7} $. The coordinates of the foci are $ (0, c) $ and $ (0, -c) $, resulting in $ (0, \sqrt{7}) $ and $ (0, -\sqrt{7}) $.

Step 5: Summarize the results. The vertices are $ (0, 4) $ and $ (0, -4) $, and the foci are $ (0, \sqrt{7}) $ and $ (0, -\sqrt{7}) $. These points describe the geometry of the ellipse.

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