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)

x264+y239=1\frac{$$x^2$$}{64}+\frac{$$y^2$$}{39}=164x2+39y2=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 noticed is that for the foci and vertices, they are both given the x coordinate here. We have negative five and five, then we have negative eight and eight. 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 squared over a squared. We have the semi major axis underneath the x, plus y squared over b squared is equal to one. 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 eight units to the right and a distance of eight units to the left. So we can say that the semi major axis a is equal to eight. 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 eight zero or negative eight zero and eight zero, and then the foci are at the points negative five zero five zero 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 two foci, which we can see based on the coordinates that we're given is five. So we're going to have that five squared is equal to a squared, which we said a was eight. So that means that we're going to have eight squared, and then this whole thing minus b squared where we just solve for b. So what I can do is recognize five squared is 25, we have eight 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 60 fours there, giving us twenty five 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 gonna 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 eight 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 one. Now at this point, I can simplify this farther by recognizing that eight 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 one. So this is the standard form equation for the ellipse that we have in this problem. So 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)

x264+y239=1\frac{$$x^2$$}{64}+\frac{$$y^2$$}{39}=164x2+39y2=1

x264+y225=1\frac{$$x^2$$}{64}+\frac{$$y^2$$}{25}=164x2+25y2=1

x225+y264=1\frac{$$x^2$$}{25}+\frac{$$y^2$$}{64}=125x2+64y2=1

x28+y25=1\frac{$$x^2$$}{8}+\frac{$$y^2$$}{5}=18x2+5y2=1

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

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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 key components of the ellipse: The foci are at (-5, 0) and (5, 0), and the vertices are at (-8, 0) and (8, 0). This indicates that the ellipse is horizontally oriented, with its center at the origin (0, 0).

Recall the standard form of the equation for a horizontally oriented ellipse:

\frac{x^{2}}{a_{2}} + \frac{y^{2}}{b_{2}} = 1

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

Determine the value of a: The distance from the center to each vertex is the semi-major axis, a. Since the vertices are at (-8, 0) and (8, 0), a = 8. Thus,