Answer:
[tex]\frac{(x-4)^2}{49}+\frac{(y-4)^2}{24}=1[/tex]
Step-by-step explanation:
Because the vertices are on the same horizontal line and so are our foci, this is our major axis for a horizontal ellipse.
Horizontal Ellipse:
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
a>b
Major axis length: 2a
Minor axis length: 2b
Center: (h,k)
Foci: (h±c,k) where a²-b²=c²
We can easily find the center of the ellipse by finding the midpoint of our vertices (-3,4) and (11,4) which works out to be (4,4)
Because our foci must be (-1,4) and (9,4) this means (4±c,4) represents our foci. We can easily see that c=5 since (4-5,4) -> (-1,4) and (4+5,4) -> (9,4).
Knowing that c=5, we need to determine our values for a and b to satisfy the equation a²-b²=5². Our vertices are located on the major axis, which will help us in finding the value of a. The distance between the two vertices will be the length of the major axis, which will be |11-(-3)| = 14. Additionally, because the length of the major axis is equal to 2a, this means 2a=14 where a=7. Now we can solve for b with our equation 7²-b²=5² where b=√24.
In conclusion, our ellipse would be:
[tex]\frac{(x-4)^2}{7^2}+\frac{(y-4)^2}{\sqrt{24}^2}=1[/tex] which works out to be [tex]\frac{(x-4)^2}{49}+\frac{(y-4)^2}{24}=1[/tex]
