Respuesta :

Hello there. To solve this question, we'll have to remember some properties about ellipses.

Given that the major axis length of the ellipse is 8 and its foci are at (-4, 4) and (-4, 0), we already have two informations that guides us into the form we're looking for.

First, the equation of an ellipse is given as

[tex]\dfrac{(x-x_0)}{a^2}+\dfrac{(y-y_0)}{b^2}=1[/tex]

Where (x0, y0) are the coordinates of the center of the ellipse, a and b are the major and minor axis (not respectively, because it depends whether the axis of symmetry of the ellipse is a x or y line).

We find the foci coordinates using the values of a and b, but in this case, as they were given, we do the opposite.

Let's start drawing the foci:

Since the foci are aligned as in a vertical line, we know that this ellipse has a major axis parallel to the y-axis.

Its center is the midpoint of the segment joining the foci, that is, the ordered pair (xm, ym) such that

[tex]\begin{gathered} x_m=\dfrac{-4+(-4)}{2}=\dfrac{-8}{2}=-4 \\ \\ y_m=\dfrac{4+0}{2}=\dfrac{4}{2}=2 \\ \end{gathered}[/tex]

Hence the center is at (-4, 2).

In the case of the major axis being parallel to the y-axis, we write the equation of the ellipse as follows:

[tex]\dfrac{(y-y_0)}{a^2}+\dfrac{(x-x_0)}{b^2}=1[/tex]

Finally, we have to determine the value of a.

We know that is the semi-major axis of the ellipse, hence it has half the length of the major axis. Since it was given to be 8, we have that

[tex]a=\dfrac{8}{2}=4[/tex]

And we determine the value of b by using the following rule for ellipses:

Therefore we get that c is the distance between the foci and the center, in this case it is simply 2.

Plugging the values we found, we get

[tex]\begin{gathered} 4^2=b^2+2^2 \\ \\ 16=b^2+4 \\ \\ b^2=12 \end{gathered}[/tex]

In this case, we can say that the semi-minor axis of this ellipse has a measure of 2sqrt(3), but as we only need the value squared, we plug it into the equation as:

[tex]\begin{gathered} \dfrac{(y-2)^2}{16}+\dfrac{(x-(-4))^2}{12}=1 \\ \\ \boxed{\dfrac{(y-2)^2}{16}+\dfrac{(x+4)^2}{12}=1} \end{gathered}[/tex]

This is the answer to this question. You can see its graph in the following image:

Ver imagen IkhlasA374544
Ver imagen IkhlasA374544
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