Question

Which points are the foci of the ellipse? (x+1)^2/225 + (y+6)^2/144 =1

Answer 1

The answer above is correct but more specific answer and on edg is (−10, −6) and (8, −6). Have a great day everyone :)

Answer 2

`\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 √( a ^2- b ^2) \end{cases}\\\\ -------------------------------`


`\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=√(15^2-12^2)\implies c=√(81)\implies c=\pm 9`

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 ).