Question Number 11338 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17 | ||
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 20/Mar/17 | ||
Answered by mrW1 last updated on 21/Mar/17 | ||
$$\left(\mathrm{1}\right) \\ $$$${let}\:{B}\left({t},\frac{\mathrm{1}}{{t}}\right) \\ $$$$\boldsymbol{{a}}=\overset{\rightarrow} {{OA}} \\ $$$$\boldsymbol{{b}}=\overset{\rightarrow} {{OB}} \\ $$$$\mathrm{cos}\:\angle{AOB}=\frac{\boldsymbol{{a}}\centerdot\boldsymbol{{b}}}{\mid\boldsymbol{{a}}\mid×\mid\boldsymbol{{b}}\mid}=\frac{{t}\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{{t}\sqrt{\mathrm{3}}}}{\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\mathrm{cos}\:\frac{\pi}{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{t}\sqrt{\mathrm{3}}+\frac{\mathrm{1}}{{t}\sqrt{\mathrm{3}}}}{\sqrt{\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{t}+\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{10}}×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{t}+\frac{\mathrm{1}}{{t}}=\frac{\sqrt{\mathrm{10}}}{\mathrm{2}}×\sqrt{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\mathrm{9}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{6}=\frac{\mathrm{10}}{\mathrm{4}}\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right) \\ $$$$\mathrm{9}{t}^{\mathrm{4}} +\mathrm{1}+\mathrm{6}{t}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{2}}\left({t}^{\mathrm{4}} +\mathrm{1}\right) \\ $$$$\mathrm{13}{t}^{\mathrm{4}} +\mathrm{12}{t}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} =\frac{−\mathrm{12}+\sqrt{\mathrm{144}+\mathrm{4}×\mathrm{13}×\mathrm{3}}}{\mathrm{2}×\mathrm{13}}=\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{13}} \\ $$$${t}=\sqrt{\frac{\mathrm{5}\sqrt{\mathrm{3}}−\mathrm{6}}{\mathrm{13}}} \\ $$ | ||
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 07/Apr/17 | ||