【答案】
分析:求出曲線方程的導(dǎo)函數(shù),根據(jù)曲線方程設(shè)出切點坐標(biāo),把設(shè)出的切點橫坐標(biāo)代入導(dǎo)函數(shù)中表示出的導(dǎo)函數(shù)值即為切線的斜率,由切點坐標(biāo)和斜率表示出切線方程,把原點坐標(biāo)代入切線方程中即可求出切點的橫坐標(biāo),進而得到切點的縱坐標(biāo)和切線的斜率.
解答:解:對
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求導(dǎo)得:y′=
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,設(shè)切點坐標(biāo)為(x
,
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),
所以切線的斜率k=
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,則切線方程為:y-
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=
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(x-x
),
把原點(0,0)代入切線方程得:x
=2,
所以切點坐標(biāo)為(2,
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),斜率為
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,
故答案為:
點評:本題的解題思想是設(shè)出切點的坐標(biāo),把切點的橫坐標(biāo)代入曲線方程的導(dǎo)函數(shù)中求出切線的斜率,進而寫出切線方程,然后把原點坐標(biāo)代入切線方程求出切點的橫坐標(biāo),從而確定出切線的斜率,屬于中檔題.