Question

10．已知質(zhì)點(diǎn)運(yùn)動(dòng)的軌跡方程為$\left\{\begin{array}{l}{x=a+tcosθ}\\{y=b+tsinθ}\end{array}\right.$（t為參數(shù)），求運(yùn)動(dòng)質(zhì)點(diǎn)從時(shí)間t₁到t₂經(jīng)過的距離．

Answer 1

分析 由已知得$\left\{\begin{array}{l}{{x}_{1}=a+{t}_{1}cosθ}\\{{y}_{1}=b={t}_{1}sinθ}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=a+{t}_{2}cosθ}\\{{y}_{2}=a+{t}_{2}sinθ}\end{array}\right.$，從而運(yùn)動(dòng)質(zhì)點(diǎn)從時(shí)間t₁到t₂經(jīng)過的距離|M₁M₂|=$\sqrt{（{x}_{1}-{x}_{2}）^{2}+（{y}_{1}-{y}_{2}）^{2}}$，由此能求出結(jié)果．

解答 解：∵質(zhì)點(diǎn)運(yùn)動(dòng)的軌跡方程為$\left\{\begin{array}{l}{x=a+tcosθ}\\{y=b+tsinθ}\end{array}\right.$（t為參數(shù)），
∴$\left\{\begin{array}{l}{{x}_{1}=a+{t}_{1}cosθ}\\{{y}_{1}=b={t}_{1}sinθ}\end{array}\right.$，$\left\{\begin{array}{l}{{x}_{2}=a+{t}_{2}cosθ}\\{{y}_{2}=a+{t}_{2}sinθ}\end{array}\right.$，
∴M₁（a+t₁cosθ，b+t₁sinθ），M₂（a+t₂cosθ，b+t₂sinθ），
∴|M₁M₂|=$\sqrt{（{x}_{1}-{x}_{2}）^{2}+（{y}_{1}-{y}_{2}）^{2}}$
=$\sqrt{（{t}_{1}-{t}_{2}）^{2}（co{s}^{2}α+si{n}^{2}α）}$
=|t₁-t₂|．
∴運(yùn)動(dòng)質(zhì)點(diǎn)從時(shí)間t₁到t₂經(jīng)過的距離為|t₁-t₂|．

點(diǎn)評(píng) 本題考查質(zhì)點(diǎn)運(yùn)動(dòng)距離的求法，是基礎(chǔ)題，解題時(shí)要認(rèn)真審題，注意兩點(diǎn)間距離公式的合理運(yùn)用．