Respuesta :
[tex]\bf \textit{ellipse, horizontal major axis}
\\\\
\cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1
\qquad
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2- b ^2}
\end{cases}\\\\
-------------------------------[/tex]
[tex]\bf \cfrac{(x+1)^2}{225}+\cfrac{(y+6)^2}{144}=1\implies \cfrac{[x-(-1)]^2}{15^2}+\cfrac{[y-(-6)]^2}{12^2}=1 \\\\\\ \begin{cases} h=-1\\ k=-6\\ a=15\\ b=12 \end{cases}\implies c=\sqrt{15^2-12^2}\implies c=\sqrt{81}\implies c=\pm 9[/tex]
since the fraction with the larger denominator is the fraction with the "x" variable, we know the ellipse is a horizontal one, with a center at (-1, -6).
since c = ±9, the foci are 9 units to the left and right of the center, thus
( -1 ± 9 , -6 ).
[tex]\bf \cfrac{(x+1)^2}{225}+\cfrac{(y+6)^2}{144}=1\implies \cfrac{[x-(-1)]^2}{15^2}+\cfrac{[y-(-6)]^2}{12^2}=1 \\\\\\ \begin{cases} h=-1\\ k=-6\\ a=15\\ b=12 \end{cases}\implies c=\sqrt{15^2-12^2}\implies c=\sqrt{81}\implies c=\pm 9[/tex]
since the fraction with the larger denominator is the fraction with the "x" variable, we know the ellipse is a horizontal one, with a center at (-1, -6).
since c = ±9, the foci are 9 units to the left and right of the center, thus
( -1 ± 9 , -6 ).
Answer:
The answer above is correct but more specific answer and on edg is (−10, −6) and (8, −6). Have a great day everyone :)
