Given:
Center ==> (4, 1)
Length = 8
c = 3
Let's find the equation for the given ellipse that statisfies the conditions above.
Take the equation:
[tex]\frac{(x-x1)^2}{a^2}+\frac{(y-y1)^2}{b^2}=1[/tex]
Here, we have a vertical minor axis with points: (x, y) ==> (4, 1) which has the equation below:
Substitute 4 for x1 and substitute 1 for y1
[tex]\frac{(x-4)^2}{a^2}+\frac{(y-1)^2}{b^2}=1[/tex]
Where:
[tex]c^2=a^2+b^2[/tex]
Since the minor axis is vertical and has a length of 8, we have:
[tex]\begin{gathered} 2b=8 \\ \\ \text{Divide both sides by 2:} \\ \frac{2b}{2}=\frac{8}{2} \\ \\ b=4 \end{gathered}[/tex]
Given: c = 3
To find the value of a, substitute 3 for c and 4 for b:
[tex]\begin{gathered} c^2=a^2+b^2^{} \\ \\ 3^2=a^2-4^2 \\ \\ \text{Add 4}^2\text{ to both sides:} \\ 3^2+4^2=a^2-4^2+4^2 \\ \\ 3^2+4^2=a^2 \\ \\ \text{Take the square root of both sides:} \\ \sqrt[]{3^2+4^2}=\sqrt[]{a^2} \\ \\ \sqrt[]{3^2+4^2}=a \\ \\ \sqrt[]{9+16}=a \\ \\ \sqrt[]{25}=a \\ \\ 5=a \\ \\ a=5 \end{gathered}[/tex]
Therefore, from the general equation, substitute 5 for a, and 4 for b
[tex]\begin{gathered} \frac{(x-4)^2}{a^2}+\frac{(y-1)^2}{b^2}=1 \\ \\ \\ \frac{(x-4)^2}{5^2}+\frac{(y-1)^2}{4^2}=1 \\ \\ \frac{(x-4)^2}{25^{}}+\frac{(y-1)^2}{16^{}}=1 \end{gathered}[/tex]
Therefore, the equation for thr given ellipse that satisfies the given conditions is:
[tex]\frac{(x-4)^2}{25^{}}+\frac{(y-1)^2}{16^{}}=1[/tex]
ANSWER:
[tex]\frac{(x-4)^2}{25^{}}+\frac{(y-1)^2}{16^{}}=1[/tex]
