Answer:
The equation of the ellipse is [tex] \frac{(x-3)^2}{169} +\frac{(y-1)^2}{144}=1[/tex]
Step-by-step explanation:
The equation of the ellipse is given by:
[tex]\frac{(x-x_0)^2}{a^2} +\frac{(y-y_0)^2}{b^2}=1[/tex]
where:
Center of the ellipse is (x₀,y₀)
Vertices: (x₀±a,y₀)
c: distance from the center to the focus [tex]c=\sqrt{a^2-b^2}[/tex]
Eccentricity [tex]e=\frac{c}{a}[/tex]
Directrix= [tex]x_0 + \frac{a^2}{c}[/tex]
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
[tex]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[/tex]
Then, we have to obtain b:
[tex]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[/tex]
The equation of the ellipse is:
[tex]\frac{(x-3)^2}{13^2} +\frac{(y-1)^2}{12^2}=1 \\ \frac{(x-3)^2}{169} +\frac{(y-1)^2}{144}=1[/tex]
Answer:
the first option is correct
Step-by-step explanation:
