Answer:
[tex]\dfrac{(x+2)^2}{16}+\dfrac{(y-1)^2}{25}=1[/tex]
Step-by-step explanation:
As the x-coordinates of the given foci are the same, the ellipse is vertical.
The general equation of a vertical ellipse is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a vertical ellipse}}\\\\\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\\\\\textsf{where:}\\\phantom{ww}\bullet\textsf{$2b=$ major axis}\\\phantom{ww}\bullet\textsf{$2a=$ minor axis}\\\phantom{ww}\bullet \textsf{$(h,k)=$ center}\\\phantom{ww}\bullet\textsf{$(h,k\pm b)=$ vertices}\\\phantom{ww}\bullet\textsf{$(h\pm a,k)=$ co-vertices}\\\phantom{ww}\bullet\textsf{$(h,k\pm c)=$ foci where $c^2=b^2-a^2$}\end{array}}[/tex]
Given that the length of the major axis is 10, then:
[tex]2b=10 \\\\ b=5 \\\\ b^2=25[/tex]
The foci of an ellipse are located at equal distances from the center (h, k) of the ellipse along the major axis. Therefore, if the foci are at (-2, 4) and (-2, -2), then:
[tex]h = -2\\\\k=\dfrac{4+(-2)}{2}=1[/tex]
So, the center of the ellipse is (-2, 1).
To find the value of c, we can use the foci formula (h, k±c):
[tex](h,k\pm c)=(-2, 1\pm c)[/tex]
Given that the y-coordinates of the foci are y = 4 and y = -2, then:
[tex]1 + c = 4\implies c = 3\\\\1 - c = -2\implies c=3[/tex]
Now, substitute the values of b and c into c² = b² - a² to find the value of a²:
[tex]3^2=5^2-a^2\\\\a^2=5^2-3^2\\\\a^2=25-9\\\\a^2=16[/tex]
Finally, substitute h = -2, k = 1, a² = 16 and b² = 25 into the given equation to write the equation of the ellipse with the given parameters:
[tex]\Large\boxed{\boxed{\dfrac{(x+2)^2}{16}+\dfrac{(y-1)^2}{25}=1}}[/tex]
