Question Number 200937 by Mingma last updated on 26/Nov/23
Answered by MM42 last updated on 26/Nov/23
$$\frac{\mathrm{25}\sqrt{\mathrm{2}}}{\mathrm{4}}\pi \\ $$
Commented by Mingma last updated on 26/Nov/23
Please show workings
Commented by Mingma last updated on 26/Nov/23
Nice solution!
Answered by witcher3 last updated on 26/Nov/23
$$\mathrm{z}=\mathrm{x}+\mathrm{iy}; \\ $$$$\Leftrightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }\leqslant\mathrm{5}\sqrt{\mathrm{2}}…..\mathrm{R} \\ $$$$\mathrm{E}=\left\{\left(\mathrm{x},\mathrm{y}\right)\mid\mathrm{R}\:\mathrm{is}\:\mathrm{True}\right\} \\ $$$$\mathrm{MF}+\mathrm{MF}'=\mathrm{2a}\:\mathrm{defind}\:\mathrm{elpise} \\ $$$$\mathrm{MF}+\mathrm{MF}'\leqslant\mathrm{2a}\:\mathrm{all}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{elipse} \\ $$$$\mathrm{2a}=\mathrm{5}\sqrt{\mathrm{2}}\Rightarrow\mathrm{a}=\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{F}\left(\mathrm{0},\mathrm{3}\right),\mathrm{F}'\left(\mathrm{4},\mathrm{0}\right) \\ $$$$\mathrm{FF}'=\mathrm{5} \\ $$$$\mathrm{c}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{b}=\sqrt{\mathrm{a}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }=\sqrt{\frac{\mathrm{25}}{\mathrm{2}}−\frac{\mathrm{25}}{\mathrm{4}}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{S}=\int\int_{\mathrm{E}} \mathrm{dxdy}=\int\int_{\mathrm{E}} \mathrm{1}_{\mathrm{E}} \mathrm{dxdy}=\mathrm{S}_{\left(\mathrm{elipse}\right)} =\pi.\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}.\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{S}_{\mathrm{elipse}} =\pi\mathrm{ab} \\ $$$$=\pi\frac{\mathrm{25}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by Mingma last updated on 26/Nov/23
Nice solution!