Find the standard form of the equation for an ellipse with the following conditions.Foci = (−5,0),(5,0)\left(-5,0\right),\left(5,0\right)(−5,0),(5,0)Vertices = (−8,0),(8,0)\left(-8,0\right),\left(8,0\right)(−8,0),(8,0)

x 2 64 + y 2 39 = 1 \frac{x^2}{64}+\frac{y^2}{39}=1 64 x 2 + 39 y 2 = 1

Find the standard form of the equation for an ellipse with the following conditions. Foci = ( − 5 , 0 ) , ( 5 , 0 ) \left(-5,0\right),\left(5,0\right) ( − 5 , 0 ) , ( 5 , 0 ) Vertices = ( − 8 , 0 ) , ( 8 , 0 ) \left(-8,0\right),\left(8,0\right) ( − 8 , 0 ) , ( 8 , 0 )

Hey, everyone. Welcome back. So let's see if we can solve this problem. So in this problem, we're told to find the standard form of the equation for an ellipse with the following conditions. And we're given that the foci are at negative five zero, positive five zero, and that the vertices are at negative eight zero, positive eight zero. Now, notice in this problem, we are given the foci and vertices, and we're asked to reverse engineer this and get the equation. So let's see what we can do. Well, what I first notice is that for the foci and vertices, they are both given the x coordinate here. We have negative 5 and 5, then we have negative 8 and 8. Since these are both the x coordinates that we see filled out, I know that this is going to be a horizontally oriented ellipse. So basically this ellipse is going to be stretched on the x axis. Now what I can do here is I can recognize that the general equation for a horizontal ellipse is going to be x x squared over a squared. We have the semi major axis underneath the x, plus y squared over b squared is equal to 1. Now, to solve this problem I really just need to find a and b, and I can do that by using the foci and the vertices. Now first off, if I look at the vertices notice that the distance to either of the vertices is going to be a distance of 8 units to the right, and a distance of 8 units to the left. So we can say that the semi major axis a is equal to 8. So pretty straightforward. Now next we need to find b, and b is a bit trickier, but we can actually do this by using the foci. Because if the vertices are at the points 80 or negative 80 and 80, and then the foci are at the points negative 50, 50 which would be right about there. Then what we can do is we can recognize that the distance c to our foci c squared is equal to a squared minus b squared. And c would be the distance to either the 2 foci, which we can see based on the coordinates that we're given is 5. So we're going to have that 5 squared is equal to a squared which we said a was 8, so that means that we're going to have 8 squared, and then this whole thing minus b squared where we just solve for b. So I can do is recognize 5 squared is 25, we have 8 squared which is 64, and this is all minus b squared, and what I can do is subtract 64 on both sides of this equation that'll cancel the 64s there, giving us 25 minuteus 64, which is negative 39. So what negative 39 is equal to negative B squared, I can cancel the negative sign on both sides and take the square root. So we're going to get that B is equal to the square root of 39. So this is our b, and that is our a value. So to find our equation for the for the ellipse that we have, we're going to have x squared over a squared, where a squared is 8 squared. We're going to have plus y squared over b squared, we said b is the square root of 39, so it's square root of 39 squared, and then this whole thing is equal to 1. Now at this point I can simplify this farther by recognizing that 8 squared is 64, and I can recognize that the square root and square are going to cancel each other giving us y squared over 39, And then this is once again all equal to 1. So this is the standard form equation for the ellipse that we have in this problem. So hope that is the answer based on the foci and vertices we were given. So hope you found this helpful. Thanks for watching.

Find the standard form of the equation for an ellipse with the following conditions. Foci = ( − 5 , 0 ) , ( 5 , 0 ) \left(-5,0\right),\left(5,0\right) ( − 5 , 0 ) , ( 5 , 0 ) Vertices = ( − 8 , 0 ) , ( 8 , 0 ) \left(-8,0\right),\left(8,0\right) ( − 8 , 0 ) , ( 8 , 0 )

x 2 64 + y 2 39 = 1 \frac{x^2}{64}+\frac{y^2}{39}=1 64 x 2 + 39 y 2 = 1

x 2 64 + y 2 25 = 1 \frac{x^2}{64}+\frac{y^2}{25}=1 64 x 2 + 25 y 2 = 1

x 2 25 + y 2 64 = 1 \frac{x^2}{25}+\frac{y^2}{64}=1 25 x 2 + 64 y 2 = 1

x 2 8 + y 2 5 = 1 \frac{x^2}{8}+\frac{y^2}{5}=1 8 x 2 + 5 y 2 = 1

8. Conic Sections

Ellipses: Standard Form

College Algebra

8. Conic Sections

Ellipses: Standard Form

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Multiple Choice

Find the standard form of the equation for an ellipse with the following conditions.
Foci = $\left(-5,0\right),\left(5,0\right)$
Vertices = $\left(-8,0\right),\left(8,0\right)$

$\frac{x^2}{64}+\frac{y^2}{25}=1$

$\frac{x^2}{25}+\frac{y^2}{64}=1$

$\frac{x^2}{8}+\frac{y^2}{5}=1$

$\frac{x^2}{64}+\frac{y^2}{39}=1$

Verified step by step guidance

Identify the center of the ellipse. Since the foci are at (-5, 0) and (5, 0), and the vertices are at (-8, 0) and (8, 0), the center is at the midpoint of the vertices, which is (0, 0).

Determine the orientation of the ellipse. The foci and vertices are along the x-axis, indicating that the major axis is horizontal.

Calculate the distance from the center to a vertex, which is the semi-major axis 'a'. The distance from (0, 0) to (8, 0) is 8, so a = 8.

Calculate the distance from the center to a focus, which is 'c'. The distance from (0, 0) to (5, 0) is 5, so c = 5.

Use the relationship c^2 = a^2 - b^2 to find the semi-minor axis 'b'. Substitute a = 8 and c = 5 into the equation: 5^2 = 8^2 - b^2, which simplifies to b^2 = 64 - 25. Solve for b^2 to find the value needed for the standard form equation.

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