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双曲线焦点三角形面积公式

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定理  在双曲线`\frac{{{x^2}}}{{{a^2}}} -\frac{{{y^2}}}{{{b^2}}} = 1``a>b>0`)中,焦点分别为`{F_1}``{F_2}`,点P是双曲线上任意一点,`\angle {F_1}P{F_2} = \theta `,则`{S_{\Delta {F_1}P{F_2}}} = {b^2}\cot \frac{\theta }{2}`.

证明:记`|P{F_1}| = {r_1},|P{F_2}| = {r_2}`,由双曲线的第一定义得

`|{r_1} - {r_2} |= 2a,\therefore {({r_1} - {r_2})^2} = 4{a^2}.`

在`△{F_1}P{F_2}`中,由余弦定理得:`{r_1}^2 + {r_2}^2 - 2{r_1}{r_2}\cos \theta  = {(2c)^2}.`

配方得:`{({r_1} - {r_2})^2} +2{r_1}{r_2} - 2{r_1}{r_2}\cos \theta  = 4{c^2}.`

即`4{a^2} +2{r_1}{r_2}(1- \cos \theta ) = 4{c^2}.`

`\therefore {r_1}{r_2} = \frac{{2({c^2} - {a^2})}}{{1 - \cos \theta }} = \frac{{2{b^2}}}{{1- \cos \theta }}.`

由任意三角形的面积公式得:

`{S_{\Delta {F_1}P{F_2}}} = \frac{1}{2}{r_1}{r_2}\sin \theta  = {b^2} \cdot \frac{{\sin \theta }}{{1 - \cos \theta }} = {b^2} \cdot \frac{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}{{2{{\sin }^2}\frac{\theta }{2}}} = {b^2} \cdot \cot \frac{\theta }{2}`.

`\therefore {S_{\Delta {F_1}P{F_2}}} = {b^2}\cot \frac{\theta }{2}.`

同理可证,在双曲线`\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1``a>b>0`)中,公式仍然成立.