已知抛物线
的焦点到准线的距离为2,过点
作抛物线
的两条切线,切点分别为
,若
,则点
到原点的距离为______ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5bf92a1ba410263d4f68b7e0432b19aa.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9e1601407780dee30f8e7e3103e4892a.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f6bce3d91ca23b86d8c6625f2632e437.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2aae6b4f34d4bad6e332031d0257ed39.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5963abe8f421bd99a2aaa94831a951e9.png)
2023·全国·模拟预测 查看更多[2]
更新时间:2023-05-04 16:28:10
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【推荐1】意大利画家列奥纳多.达·芬奇的画作《抱银貂的女人》中,女士脖颈上悬挂的黑色珍珠项链与主人相互映衬呈现出不一样的美与光泽,达·芬奇提出:固定项链的两端,使其在重力的作用下自然下垂,项链所形成的曲线是什么?这就是著名的“悬链线问题”,后人给出了悬链线的函数解析式:
,其中a为悬链线系数,
称为双曲余弦函数,其函数表达式为
,相应地双曲正弦函数的函数表达式为
.若直线
与双曲余弦函数
与双曲正弦函数
分别相交于点
、
,曲线
在点
处的切线
,曲线
在点
处的切线
相交于点
,且
为锐角三角形,则实数
的取值范围为________ .
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【推荐2】牛顿迭代法又称牛顿—拉夫逊方法,它是牛顿在17世纪提出的一种在实数集上近似求解方程根的一种方法,具体步骤如下:设
是函数
的一个零点,任意选取
作为
的初始近似值,过点
作曲线
的切线
,设
与
轴交点的横坐标为
,并称
为
的1次近似值;过点
作曲线
的切线
,设
与
轴交点的横坐标为
,称
为
的2次近似值,过点
作曲线
的切线
,记
与
轴交点的横坐标为
,并称
为
的
次近似值,设
的零点为
,取
,则
的2次近似值为__________ ;设
,数列
的前
项积为
.若任意的
恒成立,则整数
的最小值为__________ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/11bc05f41215f9894e11d1df0465751a.png)
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【推荐1】如图,两条距离为4的直线都与y轴平行,它们与抛物线
和圆
分别交于
和
,且抛物线的准线与圆相切,则当
取得最大值时,直线
的方程为_________ .
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解题方法
【推荐2】抛物线
的焦点
到准线的距离为2,过点
的直线与
交于
,
两点,
的准线与
轴的交点为
,若
的面积为
,则![](https://staticzujuan.xkw.com/quesimg/Upload/formula/38544c35c45a692f3f1c8dbcdb9db71e.png)
___________ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/fdea484053ce8c9e8edced935e593912.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c5db41a1f31d6baee7c69990811edb9f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/5963abe8f421bd99a2aaa94831a951e9.png)
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![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c5db41a1f31d6baee7c69990811edb9f.png)
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