抛物线y=x2上的任意一点到直线x-y-2=0的最短距离为（）_刷题宝

题干

抛物线y=x2上的任意一点到直线x-y-2=0的最短距离为（）

A： $\sqrt{2}$

B： $\frac{7 \sqrt{2}}{8}$

C： $2 \sqrt{2}$

D：以上答案都不对

上一题下一题 0.0难度选择题更新时间:2017-08-11 04:01:39

答案(点此获取答案解析）

B

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现代社会中人类使用的能量主要来源于（）

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如图，在△ABC中，点D、E、F分别是三条边上的点，EF∥AC，DF∥AB，∠B=45°，∠C=60°．则∠EFD=（　　）

同类题5

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{3} = 1

a > \sqrt{3}

）的右焦点为F，右顶点为A，已知

\frac{1}{| O F |} + \frac{1}{| O A |} = \frac{3 e}{| F A |}

，其中O 为原点， e为椭圆的离心率.

（Ⅰ）求椭圆的方程；

（Ⅱ）设过点A的直线l与椭圆交于点B（B不在x轴上），垂直于l的直线与l交于点M，与y轴交于点H，若 $B F ⊥ H F$ ，且 $∠ M O A = ∠ M A O$ ，求直线的l斜率.

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