Answer:
[tex]\mathsf{\dfrac{(x+4)^2}{49}+\dfrac{(y-6)^2}{40}=1}[/tex]
Step-by-step explanation:
SOLUTION:
Since the foci of the ellipse have the same y-coordinate (which is 6), the standard equation of the ellipse is:
[tex]\mathsf{\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1}[/tex]
where:
(h, k) = location of the center of the ellipse
a = semi-major axis
b = semi-minor axis
[tex]\\[/tex]
The center of the ellipse is located at the midpoint of the two foci. To solve for the location of the center, we can use the midpoint formula.
Midpoint formula:
[tex]\mathsf{x_m=\dfrac{x_1+x_2}{2}}[/tex] and [tex]\mathsf{y_m=\dfrac{y_1+y_2}{2}}[/tex]
where:
xₘ = x-coordinate of the midpoint
yₘ = y-coordinate of the midpoint
x₁ = x-coordinate of the first point
y₁ = y-coordinate of the first point
x₂ = x-coordinate of the second point
y₂ = y-coordinate of the second point
[tex]\\[/tex]
Solving for the center of the ellipse:
Let's say the first point is the first focus (-7, 6) and the second point is the second focus (-1, 6)
[tex]\mathsf{F_1=(-7,6)}[/tex]
x₁ = -7
y₁ = 6
[tex]\mathsf{F_2=(-1,6)}[/tex]
x₂ = -1
y₂ = 6
[tex]\\[/tex]
[tex]\mathsf{h=x_m=\dfrac{x_1+x_2}{2}}[/tex]
[tex]\mathsf{h=\dfrac{-7+(-1)}{2}}[/tex]
[tex]\mathsf{h=-4}[/tex]
[tex]\\[/tex]
[tex]\mathsf{k=y_m=\dfrac{y_1+y_2}{2}}[/tex]
[tex]\mathsf{k=\dfrac{6+6}{2}}[/tex]
[tex]\mathsf{k=6}[/tex]
[tex]\\[/tex]
The center of the ellipse is at (-4, 6)
[tex]\\[/tex]
The constant sum of the distances for any point from the foci, by definition, is the length of the major axis. The length of the major axis is equal to 2a.
2a = 14
[tex]\mathsf{a=\dfrac{14}{2}}[/tex]
a = 7
[tex]\\[/tex]
We already have the value of h, k and a. The only left is b. To solve for b, we know that the distance to one focus to the center of the ellipse is equal to c. If we solved to value of c we can use the relationship between a, b and c.
Relationship between a, b and c:
[tex]\mathsf{a^2=b^2+c^2}[/tex]
where:
a = semi-major axis
b = semi-minor axis
c = distance from the center to one focus
[tex]\\[/tex]
To solve for c, we can use the distance formula
Distance formula:
[tex]\mathsf{d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}[/tex]
where:
x₁ = x-coordinate of the first point
y₁ = y-coordinate of the first point
x₂ = x-coordinate of the second point
y₂ = y-coordinate of the second point
d = distance between the two points
[tex]\\[/tex]
Solving the distance between the center and one focus. Let's say the first point is our center (-4, 6) and the second point is the second focus (-1, 6).
Note: It doesn't matter what focus you use. The answer will be the same
[tex]\mathsf{C=(-4,6)}[/tex]
x₁ = -4
y₁ = 6
[tex]\mathsf{F_2=(-1,6)}[/tex]
x₂ = -1
y₂ = 6
[tex]\\[/tex]
Using the distance formula:
[tex]\mathsf{c=d=\sqrt{[-1-(-4)]^2+(6-6)^2}}[/tex]
c = 3
[tex]\\[/tex]
Using the relationship between a, b and c:
[tex]\mathsf{a^2=b^2+c^2}[/tex]
a = 7
c = 3
[tex]\\[/tex]
[tex]\mathsf{(7)^2=b^2+(3)^2}[/tex]
[tex]\mathsf{49=b^2+9}[/tex]
[tex]\mathsf{b^2=49-9}[/tex]
[tex]\mathsf{b=\sqrt{40}}[/tex]
[tex]\\[/tex]
Substituting the values of a, b, h, and k in our standard equation:
[tex]\mathsf{\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1}[/tex]
[tex]\mathsf{\dfrac{[x-(-4)]^2}{(7)^2}+\dfrac{(y-6)^2}{(\sqrt{40})^2}=1}[/tex]
[tex]\mathsf{\dfrac{(x+4)^2}{49}+\dfrac{(y-6)^2}{40}=1}[/tex] (ANSWER)
