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Szemek pisze:\(\displaystyle{ x^{2}+4y^{2}=4 \ \ \ |:{4}}\)
\(\displaystyle{ \frac{x^{2}}{4}+y^{2}=1}\)
korzystając z wzoru elipsy \(\displaystyle{ \frac{x^{2}}{a^2}+\frac{y^{2}}{b^2}=1}\)
\(\displaystyle{ \frac{x^{2}}{4}+y^{2}=1 \longrightarrow a^2=4, \ \ b^2=1}\)
Współrzędne ognisk \(\displaystyle{ F_1, \ F_2}\) elipsy:
\(\displaystyle{ F_1=(c,0), \ F_2=(-c,0)}\), gdzie \(\displaystyle{ c=\sqrt{a^2-b^2}}\)
\(\displaystyle{ c=\sqrt{4-1} \iff c=\sqrt{3}}\)
\(\displaystyle{ F_1=(\sqrt{3},0), \ F_2=(-\sqrt{3},0)}\)

\(\displaystyle{ S(0,-1)}\)
\(\displaystyle{ r=|SF_1|=|SF_2|}\)
\(\displaystyle{ r=\sqrt{(\sqrt{3}-0)^2+[0-(-1)]^2}}\)
\(\displaystyle{ |SF_1|=\sqrt{(\sqrt{3}-0)^2+[0-(-1)]^2} \iff |SF_1|=2}\)
\(\displaystyle{ r=2}\)
równanie okręgu o środku S i promieniu r:
\(\displaystyle{ o(S,r): x^2+(y+1)^2=4}\)

\(\displaystyle{ \begin{cases} x^{2}+4y^{2}=4 \\ x^2+(y+1)^2=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} x^{2}+4y^{2}=4 \\ x^2+y^2+2y+1=4 \end{cases}}\)
odejmuję drugie równanie od pierwszego
\(\displaystyle{ \begin{cases} 3y^2-2y-1=0 \ \ \ |:3 \\ x^2+y^2+2y+1=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} y^2-\frac{2}{3}y-\frac{1}{3}=0 \\ x^2+y^2+2y+1=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} (y-1)(y+\frac{1}{3})=0 \\ x^2+(y+1)^2=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x^2+(y+1)^2=4 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x^2+(y+1)^2=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x^2+(1+1)^2=4 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x^2+(-\frac{1}{3}+1)^2=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x^2+4=4 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x^2+\frac{4}{9}=4 \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x^2=0 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x^2=\frac{32}{9} \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x=0 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x=-\sqrt{\frac{32}{9}} \end{cases} \begin{cases} y=-\frac{1}{3} \\ x=\sqrt{\frac{32}{9}} \end{cases}}\)
\(\displaystyle{ \begin{cases} y=1 \\ x=0 \end{cases} \begin{cases} y=-\frac{1}{3} \\ x=-\frac{4\sqrt{2}}{3} \end{cases} \begin{cases} y=-\frac{1}{3} \\ x=\frac{4\sqrt{2}}{3} \end{cases}}\)
\(\displaystyle{ \begin{cases} x=0 \\ y=1 \end{cases} \begin{cases} x=-\frac{4\sqrt{2}}{3} \\y=-\frac{1}{3} \end{cases} \begin{cases} x=\frac{4\sqrt{2}}{3} \\ y=-\frac{1}{3} \end{cases}}\)

Niech wierzchołki trójkąta \(\displaystyle{ ABC}\) będą punktami przecięcia wykresów elipsy i okręgu.
\(\displaystyle{ A(0,1), \ \ B(-\frac{4\sqrt{2}}{3}, -\frac{1}{3}), \ \ C(\frac{4\sqrt{2}}{3},-\frac{1}{3})}\)
\(\displaystyle{ P_{\Delta{ABC}}= \frac{1}{2} ft| d(\vec{AB}, \vec{AC}) \right| = \frac{1}{2}| ft| \begin{array}{cc} x_b-x_a&y_b-y_a\\x_c-x_a&y_c-y_a \end{array} \right| | = \frac{1}{2} | ft| \begin{array}{cc} -\frac{4\sqrt{2}}{3}&-\frac{4}{3}\\\frac{4\sqrt{2}}{3}&-\frac{4}{3} \end{array} \right| | = \frac{1}{2} ft|-\frac{4\sqrt{2}}{3} ft(-\frac{4}{3}\right) - \frac{4\sqrt{2}}{3} ft(-\frac{4}{3}\right) \right|=\frac{1}{2} ft|\frac{14\sqrt{2}}{9} +\frac{14\sqrt{2}}{9}\right|=\frac{14\sqrt{2}}{9}}\)
Odpowiedź: Pole trójkąta wynosi \(\displaystyle{ \frac{14\sqrt{2}}{9}}\)