Mathematics: An ellipse has a center of (3, 1), a focus of (8, 1), and a directrix of x = 36.8. which is the equation of the ellipse?

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An ellipse has a center of (3, 1), a focus of (8,

ahmetagayew2004

08.01.2022

The equation of the ellipse is $\frac{(x-3)^2}{169} +\frac{(y-1)^2}{144}=1$

Step-by-step explanation:

The equation of the ellipse is given by:

$\frac{(x-x_0)^2}{a^2} +\frac{(y-y_0)^2}{b^2}=1$

where:

Center of the ellipse is (x₀,y₀)

Vertices: (x₀±a,y₀)

c: distance from the center to the focus $c=\sqrt{a^2-b^2}$

Eccentricity $e=\frac{c}{a}$

Directrix= $x_0 + \frac{a^2}{c}$

So we can obtain the values:

c=(8, 1)-(3, 1)=8-3= (5,0)=5

x₀= 3

y₀= 1

The directrix is x = 36.8

$36.8 = x_0 + \frac{a^2}{c}\\ 36.8 = 3 + \frac{a^2}{5}\\ 36.8 -3= \frac{a^2}{5}\\ 33.8\times 5= a^2\\ 169=a^2\\ \sqrt{169} =a\\ a=13$

Then, we have to obtain b:

$c=\sqrt{a^2-b^2}\\ c^2=a^2-b^2\\ b^2=a^2-c^2\\ b^2=13^2-5^2\\ b^2=169-25\\ b=\sqrt{144} \\ b=12$

The equation of the ellipse is:

$\frac{(x-3)^2}{13^2} +\frac{(y-1)^2}{12^2}=1 \\ \frac{(x-3)^2}{169} +\frac{(y-1)^2}{144}=1$

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