3.2.2 双曲线的简单几何性质(1)
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【基础过关系列】2022-2023学年高二数学上学期同步知识点剖析精品讲义(人教A版2019)
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基础知识
几何性质
焦点的位置 | 焦点在$x$轴上 | 焦点在$y$轴上 |
图形 | ![]() | ![]() |
标准方程 | $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a>0, b>0)$ | $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1(a>0, b>0)$ |
范围 | $x≤-a$或$x≥a$,$y∈R$ | $y≤-a$或$y≥a$,$x∈R$ |
顶点 | $A_1 (-a,0)$、$A_2 (a,0)$ | $A_1 (0,-a)$、$A_2 (0,a)$ |
轴长 | 虚轴长$2b$,实轴长$2a$ | |
焦点 | $F_1 (-c,0)$、$F_2 (c,0)$ | $F_1 (0,-c)$、$F_2 (0,c)$ |
焦距 | $F_1 F_2=2c$ | |
$a、b、c$的关系 | $c^2=b^2+a^2$ | |
离心率 | $e=\dfrac{c}{a}=\sqrt{1+\dfrac{b^2}{a^2}} \quad(e>1)$ | |
渐近线 | $y=\pm \dfrac{b}{a} x$ | $y=\pm \dfrac{a}{b} x$ |
实轴和虚轴等长的双曲线称为等轴双曲线.
【例】求双曲线 \(25x^2-16y^2=400\)的实轴长、虚轴长、离心率、焦点坐标、渐近线和顶点坐标.
解析 将方程变形为 \(\dfrac{x^2}{16}-\dfrac{y^2}{25}=1\),得\(a=4\),\(b=5\),所以 \(c=\sqrt{41}\).
故双曲线的实轴长和虚轴长分别为\(2a=8\)和\(2b=10\),离心率 \(e=\dfrac{c}{a}=\dfrac{\sqrt{41}}{4}\),
焦点坐标为 \(F_1(0,-\sqrt{41})\), \(F_2(0, \sqrt{41})\),渐近线方程为 \(y=\pm \dfrac{5}{4} x\),
顶点坐标为\(A_1 (0,-4)\),\(A_2 (0,4)\).
一些常用结论
① 通径:过焦点且垂直实轴的弦,其长度为 \(\dfrac{2 b^2}{a}\);
② 焦点到渐近线的距离是\(b\);
③ 焦点三角形面积\(S=\dfrac{b^2}{\tan \frac{\angle P}{2}}\);
④ 与双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\)共渐近线的双曲线系方程是 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=\lambda(\lambda \neq 0)\)
⑤ 焦半径\(|PF_1 |=ex_P+a\),\(|PF_2 |=ex_P-a\)(点\(P\)在双曲线右支上)
⑥ 双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\)的参数方程 \(\left\{\begin{array}{c}
x=\dfrac{a}{\cos \theta} \\
y=b \cdot \tan \theta
\end{array}\right.\) (\(θ\)为参数)
基本方法
【题型1】双曲线的简单几何性质
【典题1】已知双曲线\(C\)的方程为 \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\),则下列说法错误的是( )
A.双曲线\(C\)的实轴长为\(8\) \(\qquad \qquad \qquad \qquad\) B.双曲线\(C\)的渐近线方程为 \(y=\pm \dfrac{3}{4} x\)
C.双曲线\(C\)的焦点到渐近线的距离为\(3\) \(\qquad \qquad\) D.双曲线\(C\)上的点到焦点距离的最小值为 \(\dfrac{9}{4}\)
解析 \(∵\)双曲线\(C\)的方程为 \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\),\(∴a=4\),\(b=3\),
\(\therefore c=\sqrt{a^2+b^2}=5\),
\(∴\)实轴长为\(2a=2×4=8\),即\(A\)正确;
焦点在\(x\)轴上,则渐近线方程为 \(y=\pm \dfrac{b}{a} x=\pm \dfrac{3}{4} x\),
或解方程 \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=0,\),也可以得渐近线方程为 \(y=\pm \dfrac{3}{4} x\),即\(B\)正确;
焦点\((5,0)\)到渐近线 \(y= \dfrac{3}{4} x\)的距离为 \(\dfrac{\left|\frac{3}{4} \times 5\right|}{\sqrt{\left(\frac{3}{4}\right)^2+1}}=3\),即\(C\)正确;
对于选项\(D\),设点\(P(x,y)\)为双曲线右支上的一点,点\(F\)为双曲线的右焦点,
当\(x=4\)时,\(PF\)取最小值\(1\),即\(D\)错误.
故选:\(D\).
点拨
1.确定双曲线的几何性质,先判断焦点的位置;
2.求双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a>0, b>0)\)的渐近线,可直接解 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=0\);
3. 双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a>0, b>0)\)中焦点到渐近线的距离为虚半轴长\(b\).
【典题2】 已知双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a>0, b>0)\)的实轴长为\(4\),且两条渐近线夹角为\(60°\),则该双曲线的焦距为\(\underline{\quad \quad}\) .
解析 由题可知,\(2a=4\),\(∴a=2\),双曲线的渐近线方程为 \(y=\pm \dfrac{b}{a} x\),
\(∵\)两条渐近线的夹角为\(60°\),
\(∴\)分两种情形:①夹\(x\)轴的两条渐近线的夹角为\(60°\),
此时, \(\dfrac{b}{a}=\tan 30^{\circ}=\dfrac{\sqrt{3}}{3}\), \(\therefore b=\dfrac{2 \sqrt{3}}{3}\),
\(\therefore c=\sqrt{a^2+b^2}=\sqrt{2^2+\left(\dfrac{2 \sqrt{3}}{3}\right)^2}=\dfrac{4 \sqrt{3}}{3}\),焦距 \(2 c=\dfrac{8 \sqrt{3}}{3}\);
②夹\(y\)轴的两条渐近线的夹角为\(60°\),
此时, \(\dfrac{b}{a}=\tan 60^{\circ}=\sqrt{3}\), \(\therefore b=2 \sqrt{3}\),
\(\therefore c=\sqrt{a^2+b^2}=\sqrt{2^2+(2 \sqrt{3})^2}=4\),焦距\(2c=8\).
综上,该双曲线的焦距为\(\dfrac{8 \sqrt{3}}{3}\)或\(8\).
点拨
1.两直线的夹角\(α\),其范围为 \(\left[0, \dfrac{\pi}{2}\right]\);
2.求解双曲线的几何性质,注意对图形的分析,本题中利用了双曲线的对称性.
巩固练习
1.对于双曲线 \(C_1: \dfrac{x^2}{16}-\dfrac{y^2}{9}=1\)和 \(C_2: \dfrac{y^2}{9}-\dfrac{x^2}{16}=1\),给出下列四个结论:
(1)离心率相等;(2)渐近线相同;(3)没有公共点;(4)焦距相等,其中正确的结论是( )
A.(1)(2)(4) \(\qquad \qquad\qquad\) B.(1)(3)(4) \(\qquad \qquad\qquad\) C.(2)(3)(4) \(\qquad \qquad\qquad\) D.(2)(4)
2.已知双曲线 \(\dfrac{x^2}{5}-\dfrac{y^2}{m}=1(m>0)\)的一个焦点为\(F(-3,0)\),则其渐近线方程为\(\underline{\quad \quad}\).
3.设双曲线 \(\dfrac{x^2}{m^2+12}+\dfrac{y^2}{1-5 m}=1\)的实轴长为\(8\),则该双曲线的离心率为\(\underline{\quad \quad}\) .
参考答案
- 答案 \(C\)
解析 由题意,双曲线 \(C_1: \dfrac{x^2}{16}-\dfrac{y^2}{9}=1\)和 \(C_2: \dfrac{y^2}{9}-\dfrac{x^2}{16}=1\),
(1)离心率分别为\(\dfrac{5}{4}\), \(\dfrac{5}{3}\);(2)渐近线相同,为 \(y=\pm \dfrac{3}{4} x\);
(3)没有公共点;(4)焦距相等,为\(10\),
故选:\(C\). - 答案 \(y=\pm \dfrac{2 \sqrt{5}}{5} x\)
解析 双曲线 \(\dfrac{x^2}{5}-\dfrac{y^2}{m}=1(m>0)\)的一个焦点为\(F(-3,0)\),
可得 \(\sqrt{5+m}=3\),解得\(m=4\),
所以渐近线方程为: \(y=\pm \dfrac{2 \sqrt{5}}{5} x\). - 答案 \(\dfrac{5}{4}\)
解析 由题意,\(1-5m<0\),则 \(a^2=m^2+12\),\(b^2=5m-1\),
\(∴c^2=a^2+b^2=m^2+12+5m-1=m^2+5m+11\).
由实轴长为\(8\),得\(2a=8\),即\(a=4\).
\(∴m^2+12=16\),即\(m=2(1-5m<0)\).
\(\therefore c=\sqrt{4+10+11}=5\).
\(\therefore e=\dfrac{c}{a}=\dfrac{5}{4}\).
【题型2】利用双曲线的几何性质求双曲线方程
【典题1】 已知双曲线的一条渐近线方程为\(y=2x\),且经过点 \((4,4 \sqrt{3})\),则该双曲线的标准方程为\(\underline{\quad \quad}\) .
解析 方法1 根据题意知, \(2 \times 4>4 \sqrt{3}\),所以点 \((4,4 \sqrt{3})\)在渐近线方程\(y=2x\)的右下方,
所以该双曲线的焦点在\(x\)轴上,设标准方程为 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\),且\(a>0,b>0\);
又 \(\dfrac{b}{a}=2\),所以\(b=2a\);
又 \(\dfrac{16}{a^2}-\dfrac{48}{b^2}=1\),即 \(\dfrac{16}{a^2}-\dfrac{48}{4 a^2}=\dfrac{4}{a^2}=1\),解得 \(a^2=4\),\(b^2=16\),
所以双曲线的标准方程是 \(\dfrac{x^2}{4}-\dfrac{y^2}{16}=1\).
方法2 根据渐近线方程设双曲线的标准方程是 \(x^2-\dfrac{y^2}{4}=\lambda(\lambda \neq 0)\),代入点 \((4,4 \sqrt{3})\),
计算得 \(k=16-\dfrac{48}{4}=4\),
所以双曲线的标准方程为 \(x^2-\dfrac{y^2}{4}=4\),即 \(\dfrac{x^2}{4}-\dfrac{y^2}{16}=1\).
点拨
1.求双曲线方程的方法主要是待定系数法,注意焦点的位置;
2.与双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1\)共渐近线的双曲线系方程是 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=\lambda(\lambda \neq 0)\).
巩固练习
1.设椭圆\(C_1\)的离心率为\(\dfrac{5}{13}\),焦点在\(x\)轴上且长轴长为\(26\),若曲线\(C_2\)上的点到椭圆\(C_1\)的两焦点的距离差的绝对值等于\(8\),则曲线\(C_2\)的标准方程为\(\underline{\quad \quad}\) .
2.与双曲线 \(C: \dfrac{x^2}{2}-y^2=1\)共渐近线,且经过 \(\left(3, \dfrac{\sqrt{10}}{2}\right)\)点的双曲线的标准方程是\(\underline{\quad \quad}\).
参考答案
- 答案 \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\)
解析 在椭圆\(C_1\)中,由 \(\left\{\begin{array}{l} 2 a=26 \\ \dfrac{c}{a}=\dfrac{5}{13} \end{array}\right.\)得 \(\left\{\begin{array}{l} a=13 \\ c=5 \end{array}\right.\),焦点\(F_1 (-5,0)\),\(F_2 (5,0)\),
曲线\(C_2\)是以\(F_1\),\(F_2\)为焦点,实轴长为\(8\)的双曲线,
故\(C_2\)的标准方程为 \(\dfrac{x^2}{16}-\dfrac{y^2}{9}=1\). - 答案 \(\dfrac{x^2}{4}-\dfrac{y^2}{2}=1\)
解析 根据题意,要求双曲线与双曲线 \(C: \dfrac{x^2}{2}-y^2=1\)共渐近线,
设要求的双曲线为 \(\dfrac{x^2}{2}-y^2=t,(t \neq 0)\),
又由双曲线经过点 \(\left(3, \dfrac{\sqrt{10}}{2}\right)\),
则有 \(\dfrac{9}{2}-\dfrac{10}{4}=t\),解可得\(t=2\),
则要求双曲线的标准方程为 \(\dfrac{x^2}{4}-\dfrac{y^2}{2}=1\).
【题型3】双曲线的几何性质综合问题
【典题1】 已知双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左焦点为\(F_1\),右顶点为\(D\),过点\(D\)作垂直于\(x\)轴的直线交双曲线的两条渐近线于\(A\)、\(B\)两点,\(△ABF_1\)为等边三角形,则双曲线离心率为( )
A. \(\sqrt{3}\) \(\qquad \qquad\) B. \(2 \sqrt{2}\) \(\qquad \qquad\) C.\(2\) \(\qquad \qquad\) D.\(3\)
解析 如图,
双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的一条渐近线方程为 \(y=\dfrac{b}{a} x\),
取\(x=a\),得\(y=b\),则\(B(a,b)\),
由\(△ABF_1\)为等边三角形,得 \(\tan 30^{\circ}=\dfrac{b}{a+c}=\dfrac{\sqrt{3}}{3}\),
\(∴3b^2=(a+c)^2\),即 \(3(c^2-a^2 )=a^2+2ac+c^2\),
整理得\(e^2-e-2=0\),解得\(e=-1\)(舍)或\(e=2\).
故选:\(C\).
【典题2】设双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点分别为\(F_1\),\(F_2\),\(|F_1 F_2 |=2c\),过\(F_2\)作\(x\)轴的垂线,与双曲线在一第象限的交点为\(A\),点\(Q\)坐标为 \(\left(c, \dfrac{3 a}{2}\right)\)且满足\(|F_2 Q|>|F_2 A|\),若在双曲线\(C\)的右支上存在点\(P\)使得 \(\left|P F_1\right|+|P Q|<\dfrac{7}{6}\left|F_1 F_2\right|\)成立,则双曲线\(C\)的离心率的取值范围是\(\underline{\quad \quad}\) .
解析 令\(x=c\)代入双曲线的方程可得 \(y=\pm b \sqrt{\dfrac{c^2}{a^2}-1}=\pm \dfrac{b^2}{a}\),
由\(|F_2 Q|>|F_2 A|\),可得 \(\dfrac{3 a}{2}>\dfrac{b^2}{a}\),
即为 \(3a^2>2b^2=2(c^2-a^2)\),即有 \(e=\dfrac{c}{a}<\dfrac{\sqrt{10}}{2}\),①
又 \(\left|P F_1\right|+|P Q|<\dfrac{7}{6}\left|F_1 F_2\right|\)恒成立,
由双曲线的定义,可得 \(2 a+\left|P F_2\right|+|P Q|<\dfrac{7}{3} c \mid\)恒成立,
由\(F_2,P,Q\)共线时,\(|PF_2 |+|PQ|\)取得最小值 \(\left|F_2 Q\right|=\dfrac{3}{2} a\),
可得 \(\dfrac{7}{3} c>2 a+\dfrac{3}{2} a\),即有 \(e=\dfrac{c}{a}>\dfrac{3}{2}\),②
由\(e>1\),结合①②可得,e\(e>1\)的范围是 \(\left(\dfrac{3}{2}, \dfrac{\sqrt{10}}{2}\right)\).
点拨 求双曲线离心率 \(e=\dfrac{c}{a}\)值(或求离心率的取值范围),由于有 \(c^2=b^2+a^2\),则只需要得到\(a,b,c\)中任意两个参数间的齐次方程(或不等式)便可.
【典题3】已知双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的一条渐近线方程为 \(y=\dfrac{\sqrt{3}}{2} x\),\(P\)为双曲线上一个动点,\(F_1\) 、\(F_2\)为其左,右焦点, \(\overrightarrow{P F}_1 \cdot \overrightarrow{P F}_2\)的最小值为\(-3\),则此双曲线的焦距为\(\underline{\quad \quad}\).
解析 双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的一条渐近线方程为 \(y=\dfrac{\sqrt{3}}{2} x\), \(\therefore \dfrac{b}{a}=\dfrac{\sqrt{3}}{2}\),
不妨设\(a=2k\), \(b=\sqrt{3} k\),\(k>0\),
\(\therefore c=\sqrt{a^2+b^2}=\sqrt{7} k\), \(\therefore F_1(-\sqrt{7} k, 0), F_2(\sqrt{7} k, 0)\),
设\(P(x_0,y_0)\),且\(x_0≤-2k\)或\(x_0≥2k\),即 \(x_0^2 \geq 4 k^2\),
\(\therefore \dfrac{x_0^2}{a^2}-\dfrac{y_0^2}{b^2}=1\), \(\therefore y_0^2=\dfrac{3}{4} x_0^2-3 k^2\),
\(\therefore \overrightarrow{P F}_1 \cdot \overrightarrow{P F}_2=\left(-\sqrt{7} k-x_0,-y_0\right)\left(\sqrt{7} k-x_0,-y_0\right)=x_0^2-7 k^2+y_0^2\)
\(=\dfrac{7}{4} x_0^2-10 k^2\geq 7 k^2-10 k^2=-3 k^2=-3\),
解得\(k=1\)或\(k=-1\)(舍去),
\(\therefore c=\sqrt{7}\), \(\therefore 2 c=2 \sqrt{7}\),
点拨 利用设元,用坐标表示向量从而把向量相关问题转化为代数问题处理.
巩固练习
1.在平面直角坐标系\(xOy\)中,已知双曲线 \(\dfrac{x^2}{3}-\dfrac{y^2}{b^2}=1\)的两条渐近线与直线 \(x=\sqrt{3}\)围成正三角形,则双曲线的离心率为\(\underline{\quad \quad}\).
2.过双曲线 \(\dfrac{x^2}{3}-y^2=1\)的右焦点\(F\),作倾斜角为\(60°\)的直线\(l\),交双曲线的渐近线于点\(A\)、\(B\),\(O\)为坐标原点,则\(△OAB\)的面积为\(\underline{\quad \quad}\) .
3.已知双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左右焦点为\(F_1 (-2,0)\),\(F_2 (2,0)\),点\(P\)是双曲线上任意一点,若 \(\overrightarrow{P F}_1 \cdot \overrightarrow{P F}_2\)的最小值是\(-2\),则双曲线\(C\)的离心率为\(\underline{\quad \quad}\).
4.设双曲线 \(\dfrac{x^2}{16}-\dfrac{y^2}{12}=1\)的左、右焦点分别为\(F_1\),\(F_2\),过\(F_1\)的直线l交双曲线左支于\(A\),\(B\)两点,则\(|AF_2 |+|BF_2 |\)的最小值为\(\underline{\quad \quad}\).
5.已知点\(F_1 (-3,0)\),\(F_2 (3,0)\)分别是双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点,\(M\)是\(C\)右支上的一点,\(MF_1\)与\(y\)轴交于点\(P\),\(△MPF_2\)的内切圆在边\(PF_2\)上的切点为\(Q\),若\(|PQ|=2\),则\(C\)的离心率为\(\underline{\quad \quad}\) .
参考答案
-
答案 \(\dfrac{2 \sqrt{3}}{3}\)
解析 双曲线\(\dfrac{x^2}{3}-\dfrac{y^2}{b^2}=1\)的两条渐近线与直线\(x=\sqrt{3}\)围成正三角形,
所以双曲线的渐近线的倾斜角为\(30°\)和\(150°\),
所以 \(\dfrac{b}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\),所以\(b=1\),
所以双曲线的离心率为: \(e=\dfrac{c}{a}=\dfrac{2}{\sqrt{3}}=\dfrac{2 \sqrt{3}}{3}\). -
答案 \(\dfrac{3 \sqrt{3}}{2}\)
解析 不妨设点\(A\)在第一象限,点\(B\)在第四象限,
因为\(∠OFB=60°\),双曲线 \(\dfrac{x^2}{3}-y^2=1\)的渐近线方程: \(y=\pm \dfrac{\sqrt{3}}{3} x\),
所以\(∠AOF=30°\),所以\(∠FOB=30°\),
所以\(∠OBA=∠OBF=90°\),所以 \(|O B|=|O F| \cos 30^{\circ}=\sqrt{3}\).
又\(∠AOB=60°\),则\(∠OAB=30°\),
所以 \(|O A|=2|O B|=2 \sqrt{3}\),所以\(|AB|=3\),
从而\(△OAB\)的面积为: \(\dfrac{1}{2} \cdot|O A||O B| \sin 60^{\circ}=\dfrac{3 \sqrt{3}}{2} .\)
-
答案 \(\sqrt{2}\)
解析 设\(P(x_0,y_0)\),则 \(\dfrac{x_0^2}{a^2}-\dfrac{y_0{ }^2}{b^2}=1 \Rightarrow x_0^2=a^2+\dfrac{a^2}{b^2} y_0^2\).
\(∵F_1 (-2,0)\),\(F_2 (2,0)\),
\(∴c^2=4=a^2+b^2\),
\(\overrightarrow{P F}_1 \cdot \overrightarrow{P F}_2=\left(x_0+2\right)\left(x_0-2\right)+y_0^2=x_0^2+y_0^2-4\)\(=\dfrac{c^2}{b^2} y_0^2+a^2-4 \geq a^2-4\),
\(\because \overrightarrow{P F}_1 \cdot \overrightarrow{P F}_2\)的最小值是\(-2\),
\(∴a^2-4=-2\),解得 \(a=\sqrt{2}\),又\(c=2\),\(∴\)离心率\(e=\sqrt{2}\).
故答案为: \(\sqrt{2}\). -
答案 \(22\)
解析 根据双曲线 \(\dfrac{x^2}{16}-\dfrac{y^2}{12}=1\),得\(a=4\), \(b=2 \sqrt{3}\),
由双曲线的定义可得:\(|AF_2 |-|AF_1 |=2a=8\)…①,
\(|BF_2 |-|BF_1 |=2a=8\)…②,
①+②可得:\(|AF_2 |+|BF_2 |-(|AF_1 |+|BF_1 |)=16\),
由于过双曲线的左焦点\(F_1\)的直线交双曲线的左支于\(A\),\(B\)两点,
可得\(|AF_1 |+|BF_1 |=|AB|\),当\(|AB|\)是双曲线的通径时\(|AB|\)最小.
即有\(|AF_2 |+|BF_2 |-(|AF_1 |+|BF_1 |)=|AF_2 |+|BF_2 |-|AB|=16\).
即有 \(\left|B F_2\right|+\left|A F_2\right|=|A B|+16 \geq \dfrac{2 b^2}{a}+16=\dfrac{2 \times 12}{4}+16=22\). -
答案 \(\dfrac{3}{2}\)
解析 设\(△MPF_2\)的内切圆与\(MF_1\),\(MF_2\)的切点分别为\(A\),\(B\),
由切线长定理可知\(MA=MB\),\(PA=PQ\),\(BF_2=QF_2\),
又\(PF_1=PF_2\),
\(∴MF_1-MF_2=(MA+AP+PF_1 )-(MB+BF_2 )=PQ+PF_2-QF_2=2PQ\),
由双曲线的定义可知\(MF_1-MF_2=2a\),
故而\(a=PQ=2\),又\(c=3\),
\(∴\)双曲线的离心率为 \(e=\dfrac{c}{a}=\dfrac{3}{2}\).
分层练习
【A组---基础题】
1.已知双曲线的方程为 \(\dfrac{y^2}{4}-\dfrac{x^2}{9}=1\),则下列关于双曲线说法正确的是( )
A.虚轴长为\(4\) \(\qquad \qquad\) B.焦距为 \(12 \sqrt{5}\) \(\qquad \qquad\) C.离心率为 \(\dfrac{\sqrt{23}}{3}\) \(\qquad \qquad\) D.渐近线方程为\(2x±3y=0\)
2.双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点分别为\(F_1\),\(F_2\),过\(F_2\)且垂直于\(x\)轴的直线与双曲线\(C\)的两条渐近线分别交于\(M\),\(N\)两点,若\(△MF_1 N\)为等腰直角三角形,则该双曲线离心率为( )
A. \(\sqrt{15}\) \(\qquad \qquad\) B. \(\dfrac{\sqrt{15}}{2}\) \(\qquad \qquad\) C. \(\dfrac{\sqrt{5}}{3}\) \(\qquad \qquad\) D. \(\sqrt{5}\)
3.(多选)若双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的实轴长为\(6\),焦距为\(10\),右焦点为\(F\),则下列结论正确的是( )
A.\(C\)的渐近线上的点到\(F\)距离的最小值为\(4\)\(\qquad \qquad\) B.\(C\)的离心率为 \(\dfrac{5}{4}\)
C.\(C\)上的点到F距离的最小值为\(2\) \(\qquad \qquad \qquad\) D.过\(F\)的最短的弦长为 \(\dfrac{32}{3}\)
4.(多选)已知双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的离心率为 \(\dfrac{2 \sqrt{3}}{3}\),右顶点为\(A\),以\(A\)为圆心,\(b\)为半径作圆\(A\),圆\(A\)与双曲线\(C\)的一条渐近线交于\(M\)、\(N\)两点,则有 ( )
A.渐近线方程为 \(y=\pm \sqrt{3} x\) \(\qquad\) B.渐近线方程为 \(y=\pm \dfrac{\sqrt{3}}{3} x\) \(\qquad\) C.\(∠MAN=60^°\) \(\qquad\) D.\(∠MAN=120^°\)
5.(多选)已知\(P\)为双曲线 \(C: \dfrac{x^2}{3}-y^2=1\)上的动点,过\(P\)作两渐近线的垂线,垂足分别为\(A\),\(B\),记线段\(PA\),\(PB\)的长分别为\(m\),\(n\),则( )
A.若\(PA\),\(PB\)的斜率分别为\(k_1\),\(k_2\),则\(k_1⋅k_2=-3\)
B. \(m n=\dfrac{1}{2}\)
C.\(4m+n\)的最小值为 \(\sqrt{3}\)
D.\(|AB|\)的最小值为 \(\dfrac{3}{2}\)
6.焦点在\(y\)轴上,虚轴长为\(8\),焦距为\(10\)的双曲线的标准方程是\(\underline{\quad \quad}\).
7.已知 \(F(-\sqrt{3}, 0)\)是双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左焦点,\(P\)为双曲线\(C\)右支上一点,圆 \(x^2+y^2=a^2\)与\(y\)轴的正半轴交点为\(A\),\(|PA|+|PF|\)的最小值\(4\),则双曲线\(C\)的实轴长为\(\underline{\quad \quad}\) .
8.已知双曲线 \(\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点分别\(F_1\),\(F_2\),过\(F_2\)的直线交双曲线右支于\(A\),\(B\)两点.\(∠F_1 AF_2\)的平分线交\(BF_1\)于\(D\),若 \(\overrightarrow{A D}=\dfrac{1}{2} \overrightarrow{A F_1}+\overrightarrow{A F_2}\),则双曲线的离心率为\(\underline{\quad \quad}\).
9.双曲线 \(C: \dfrac{x^2}{4}-y^2=1\)的左,右顶点分别是\(A_1\),\(A_2\),\(P\)是\(C\)上任意一点,直线\(PA_1\),\(PA_2\)分别与直线\(l:x=1\)交于\(M\),\(N\),则\(|MN|\)的最小值是\(\underline{\quad \quad}\).
参考答案
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答案 \(D\)
解析 双曲线的方程为 \(\dfrac{y^2}{4}-\dfrac{x^2}{9}=1\),可得虚轴长为\(6\),实轴长为\(4\),
离心率 \(e=\dfrac{\sqrt{13}}{2}\),渐近线方程为:\(2x±3y=0\).
故选:\(D\). -
答案 \(D\)
解析 \(\left\{\begin{array}{l} x=c \\ y=\dfrac{b}{a} x \end{array}\right.\)解得 \(M\left(c, \dfrac{b c}{a}\right)\),因为\(△MF_1 N\)为等腰直角三角形,所以 \(\dfrac{b c}{a}=2 c\), \(\dfrac{b}{a}=2\),
所以 \(e=\dfrac{c}{a}=\sqrt{1+\left(\dfrac{b}{a}\right)^2}=\sqrt{5}\),故选:\(D\). -
答案 \(AC\)
解析 由题意可得\(2a=6\),\(2c=10\),
所以\(a=3\),\(c=5\), \(b=\sqrt{c^2-a^2}=4\),右焦点\(F(5,0)\),渐近线的方程为\(4x-3y=0\),
所以\(C\)的渐近线上的点到\(F\)距离的最小值\(F\)到渐近线的距离 \(d=\dfrac{b c}{c}=b=4\),所以\(A\)正确,
离心率 \(e=\dfrac{c}{a}=\dfrac{5}{3}\),所以\(B\)不正确;
双曲线上的点为顶点到焦点的距离最小,\(5-3=2\),所以\(C\)正确;
过焦点的弦长为垂直与\(x\)轴的直线与双曲线的弦长, \(\dfrac{2 b^2}{a}=\dfrac{32}{3}\),而斜率为\(0\)时,弦长为实轴长 \(2 a=6<\dfrac{32}{3}\),所以最短的弦长为\(6\),故\(D\)不正确,
故选:\(AC\). -
答案 \(BC\)
解析 由题意可得 \(e=\dfrac{c}{a}=\dfrac{2 \sqrt{3}}{3}\),
可设\(c=2t\), \(a=\sqrt{3} t\),\(t>0\),则 \(b=\sqrt{c^2-a^2}=t\), \(A(\sqrt{3} t, 0)\),
圆\(A\)的圆心为 \(A(\sqrt{3} t, 0)\),半径\(r\)为\(t\),
双曲线的渐近线方程为 \(y=\pm \dfrac{b}{a} x\),即 \(y=\pm \dfrac{\sqrt{3}}{3} x\),
圆心\(A\)到渐近线的距离为 \(d=\dfrac{\left|\frac{\sqrt{3}}{3} \cdot \sqrt{3} t\right|}{\sqrt{1+\frac{1}{3}}}=\dfrac{\sqrt{3}}{2} t\),
弦长 \(|M N|=2 \sqrt{r^2-d^2}=2 \sqrt{t^2-\dfrac{3}{4} t^2}=t=b\),
可得三角形\(MNA\)为等边三角形,即有\(∠MAN=60°\).
故选:\(BC\). -
答案 \(AD\)
解析 如右图所示,设\(P(x_0,y_0)\),则 \(\dfrac{x_0^2}{3}-y_0^2=1\).由题设条件知:
双曲线\(C\)的两渐近线: \(l_1: y=\dfrac{\sqrt{3}}{3} x, l_2: y=-\dfrac{\sqrt{3}}{3} x\).
设直线\(PA\),\(PB\)的斜率分别为\(k_1\),\(k_2\),则 \(k_1=-\sqrt{3}\), \(k_2=\sqrt{3}\),
所以\(k_1⋅k_2=-3\),故\(A\)选项正确;
由点线距离公式知: \(|P A|=m=\dfrac{\left|\sqrt{3} x_0-3 y_0\right|}{2 \sqrt{3}}\), \(|P B|=n=\dfrac{\left|\sqrt{3} x_0+3 y_0\right|}{2 \sqrt{3}}\),
\(\therefore m n=\dfrac{\left|3 x_0{ }^2-9 y_0{ }^2\right|}{12}=\dfrac{9}{12} \times\left|\dfrac{x_0{ }^2}{3}-y_0{ }^2\right|=\dfrac{3}{4}\),故\(B\)选项错误;
\(\because 4 m+n \geq 4 \sqrt{n m}=4 \times \dfrac{\sqrt{3}}{2}=2 \sqrt{3}\),所以\(C\)不正确;
由四边形\(AOBP\)中,所以\(∠APB=120^°\),
\(|A B|=\sqrt{P A^2+P B^2-2 P A \cdot P B \cdot \cos \angle A P B}\)\(=\sqrt{m^2+n^2-2 m n\left(-\dfrac{1}{2}\right)} \geq \sqrt{3 m n}=\dfrac{3}{2}\),
所以\(D\)正确,
故选:\(AD\).
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答案 \(\dfrac{y^2}{9}-\dfrac{x^2}{16}=1\)
解析 由题意,设方程为 \(\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1(a>0, b>0)\),
\(∵\)虚轴长为\(8\),焦距为\(10\),\(∴b=4\), \(a=\sqrt{5^2-4^2}=3\),
\(∴\)双曲线的标准方程是 \(\dfrac{y^2}{9}-\dfrac{x^2}{16}=1\). -
答案 \(2\)
解析 由题意,\(A(0,a)\),
设\(F'\)为双曲线的右焦点,则\(|PF|=2a+|PF'|\), \(F(-\sqrt{3}, 0)\), \(F^{\prime}(\sqrt{3},0)\).
\(\therefore|P A|+|P F|=|P A|+2 a+\left|P F^{\prime}\right|=2 a+\left(|P A|+\left|P F^{\prime}\right|\right)\)\(\geq 2 a+\left|A F^{\prime}\right|=2 a+\sqrt{3+a^2}\),
三点\(P,A,F'\)共线时取等号.
所以 \(2 a+\sqrt{3+a^2}=4\),解得\(a=1\),故实轴长为\(2\). -
答案 \(\sqrt{3}\)
解析 如图,取\(AF_1\)的中点\(E\),连接\(DE\),\(DF_2\),
由 \(\overrightarrow{A D}=\dfrac{1}{2} \overrightarrow{A F_1}+\overrightarrow{A F_2}\),可知四边形\(AF_2 DE\)为平行四边形,
又\(∵AD\)为\(∠F_1 AF_2\)的平分线,\(∴\)四边形\(AF_2 DE\)为菱形.
\(∵DE∥AB\),\(∴D\)为\(BF_1\)的中点,
\(∵DF_2∥AF_1\),\(∴F_2\)为\(AB\)的中点,则\(AE=EF_1=F_2 A=2a\).
由双曲线的对称性可知,\(AB⊥x\)轴,
\(\therefore\left|F_1 F_2\right|^2=\left|A F_1\right|^2-\left|A F_2\right|^2\),即 \(4 c^2=(4 a)^2-(2 a)^2\),解得: \(e=\dfrac{c}{a}=\sqrt{3}\).
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答案 \(\sqrt{3}\)
解析 由双曲线的对称性可知,\(P\)在右支上时,\(|MN|\)取最小值.
由上可得\(A_1 (-2,0)\),\(A_2 (2,0)\),根据双曲线方程 \(C: \dfrac{x^2}{4}-y^2=1\)可得 \(\dfrac{y}{x-2} \cdot \dfrac{y}{x+2}=\dfrac{1}{4}\),
所以设直线\(PA_1\) 、\(PA_2\)的斜率分别为\(k_1 、k_2 (k_1 、k_2>0)\),
则 \(k_1 k_2=\dfrac{1}{4}\).
\(PA_1\)的方程为\(y=k_1 (x+2)\),令\(x=1\),解得\(M(1,3k_1)\),
\(PA_2\)的方程为\(y=k_2 (x-2)\),令\(x=1\),解得\(N(1,-k_2)\),
所以 \(| M N|=| 3 k_1-\left(-k_2\right) \mid=3 k_1+k_2 \geq 2 \sqrt{3 k_1 k_2}=\sqrt{3}\),
(当且仅当\(3k_1=k_2\),即 \(k_1=\dfrac{\sqrt{3}}{6}, \quad k_2=\dfrac{\sqrt{3}}{2}\)时等号成立).
【B组---提高题】
1.已知双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点分别为\(F_1\),\(F_2\),过\(F_1\)作斜率为 \(\dfrac{\sqrt{2}}{2}\)的直线\(l\)与双曲线\(C\)的左、右两支分别交于\(A\)、\(B\)两点,若\(|AF_2 |=|BF_2 |\),则双曲线的离心率为( )
A.\(2\) \(\qquad \qquad\) B. \(\sqrt{2}\) \(\qquad \qquad\)C. \(\sqrt{5}\) \(\qquad \qquad\) D. \(\sqrt{3}\)
2.(多选)已知双曲线\(C\)的左、右焦点分别为\(F_1\),\(F_2\),过\(F_2\)的直线与双曲线的右支交于\(A\)、\(B\)两点,若\(|AF_1 |=|BF_2 |=2|AF_2 |\),则( )
A.\(∠AF_1 B=∠F_1 AB\) \(\qquad \qquad\) \(\qquad \qquad \qquad\) B.双曲线的离心率 \(e=\dfrac{\sqrt{33}}{3}\)
C.双曲线的渐近线方程为 \(y=\pm \dfrac{2 \sqrt{6}}{3} x\) \(\qquad \qquad\) D.原点\(O\)在以\(F_2\)为圆心,\(AF_2\)为半径的圆上
3.设点\(F_1\),\(F_2\)分别为双曲线 \(C: \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1(a, b>0)\)的左、右焦点,点\(A\)、\(B\)分别在双曲线\(C\)的左、右支上,若 \(\overrightarrow{F_1 B}=6 \overrightarrow{F_1 A}\), \(\overrightarrow{A F}_2^2=\overrightarrow{A B} \cdot \overrightarrow{A F_2}\),且 \(\left|\overrightarrow{A F_2}\right|<\left|\overrightarrow{B F_2}\right|\),则双曲线\(C\)的渐近线方程为\(\underline{\quad \quad}\) .
参考答案
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答案 \(D\)
解析 如图,取\(AB\)中点\(M\),连结\(F_2 M\),
\(∵|AF_2 |=|BF_2 |\),\(∴F_2M⊥AB\),
设\(|AF_2 |=|BF_2 |=x\),\(∵|AF_2 |-|AF_1 |=2a\),\(∴|AF_1 |=x-2a\),
又\(|BF_1 |-|BF_2 |=2a\),\(∴|BF_1 |=x+2a\),
\(∴|AB|=|BF_1 |-|AF_1 |=4a\),\(∴|AM|=|BM|=2a\),
\(∴|F_1 M|=|BF_1 |-|BM|=x\),
由勾股定理,知 \(\left|F_2 M\right|=\sqrt{\left(F_1 F_2\right)^2-\left(M F_1\right)^2}=\sqrt{\left(B F_2\right)^2-(B M)^2}\),
即\(\left|F_2 M\right|=\sqrt{4 c^2-x^2}=\sqrt{x^2-4 a^2}\),解得 \(x^2=2a^2+2c^2\),
\(\therefore\left|F_2 M\right|=\sqrt{2 c^2-2 a^2}=\sqrt{2 b^2}\),
\(\therefore \tan \angle M F_1 F_2=\dfrac{\left|F_2 M\right|}{\left|F_1 M\right|}=\dfrac{\sqrt{2 b^2}}{\sqrt{2 a^2+2 c^2}}=\dfrac{\sqrt{2}}{2}\),即 \(\dfrac{c^2-a^2}{a^2+c^2}=\dfrac{1}{2}\),化简得 \(c^2=3a^2\),
\(∴\)离心率 \(e=\dfrac{c}{a}=\sqrt{3}\).
故选:\(D\). -
答案 \(ABC\)
解析 根据题意,作图如下,
设\(|AF_1 |=|BF_2 |=2|AF_2 |=2m\),则\(|AB|=|AF_2 |+|BF_2 |=3m\),
由双曲线的定义知,\(|AF_1 |-|AF_2 |=2m-m=2a\),即\(m=2a\);
\(|BF_1 |-|BF_2 |=2a\),即\(|BF_1 |-2m=2a\),
\(∴|BF_1 |=3m=|AB|\),\(∴∠AF_1 B=∠F_1 AB\),即选项\(A\)正确;
由余弦定理知,在\(△ABF_1\)中, \(\cos \angle A F_1 B=\dfrac{\left|A F_1\right|^2+\left|B F_1\right|^2-|A B|^2}{2 \cdot\left|A F_1\right| \cdot\left|B F_1\right|}=\dfrac{4 m^2+9 m^2-9 m^2}{2 \cdot 2 m \cdot 3 m}=\dfrac{1}{3}\),
在\(△AF_1 F_2\)中, \(\cos \angle F_1 A B=\dfrac{\left|A F_1\right|^2+\left|A F_2\right|^2-\left|F_1 F_2\right|^2}{2 \cdot\left|A F_1\right| \cdot\left|A F_2\right|}=\dfrac{4 m^2+m^2-4 c^2}{2 \cdot 2 m \cdot m}=\cos \angle A F_1 B=\dfrac{1}{3},\),
化简整理得, \(12c^2=11m^2=44a^2\),
\(∴\)离心率 \(e=\dfrac{c}{a}=\sqrt{\dfrac{44}{12}}=\dfrac{\sqrt{33}}{3}\),即选项\(B\)正确;
双曲线的渐近线方程为 \(y=\pm \dfrac{b}{a} x=\pm \sqrt{\dfrac{c^2-a^2}{a^2}} x=\pm \sqrt{e^2-1} x=\pm \dfrac{2 \sqrt{6}}{3} x\),即选项\(C\)正确;
若原点\(O\)在以\(F_2\)为圆心,\(AF_2\)为半径的圆上,则\(c=m=2a\),与 \(\dfrac{c}{a}=\dfrac{\sqrt{33}}{3}\)不符,
故选项\(D\)错误.
故选:\(ABC\). -
答案 \(y=\pm \dfrac{2 \sqrt{10}}{5} x\)
解析 \(\because \overrightarrow{F_1 B}=6 \overrightarrow{F_1 A}\),\(∴F_1,A,B\)共线,且 \(|\overrightarrow{A B}|=5\left|\overrightarrow{A F_1}\right|\),
\(\because \overrightarrow{A F}_2^2=\overrightarrow{A B} \cdot \overrightarrow{A F_2}=\left(\overrightarrow{A F}_2+\overrightarrow{F_2 B}\right) \cdot \overrightarrow{A F_2}=\overrightarrow{A F}_2^2+\overrightarrow{F_2 B} \cdot \overrightarrow{A F_2}\),
\(\therefore \overrightarrow{F_2 B} \cdot \overrightarrow{A F_2}=0\),则 \(\overrightarrow{F_2B} \perp \overrightarrow{A F_2}\).故有 \(\left|\overrightarrow{A F_2}\right|^2+\left|\overrightarrow{B F_2}\right|^2=|\overrightarrow{A B}|^2\),
设 \(\left|\overrightarrow{A F_1}\right|=m\),则 \(|\overrightarrow{A B}|=5 m\),\(\left|\overrightarrow{B F_1}\right|=6 \mathrm{~m}\)
由双曲线的定义可得 \(\left|\overrightarrow{A F_2}\right|-m=2 a\),\(6 m-\left|\overrightarrow{B F_2}\right|=2 a\)且有 \(\left|\overrightarrow{A F}_2\right|^2+\left|\overrightarrow{B F_2}\right|^2=25 m^2\),
解得\(m=a\)或 \(m=\dfrac{2}{3} a\).
若\(m= a\),则 \(\left|\overrightarrow{A F}_2\right|=3 a<\left|\overrightarrow{B F}_2\right|=4 a\),舍去;
若 \(m=\dfrac{2}{3} a\),则 \(\left|\overrightarrow{A F}_2\right|=\dfrac{8}{3} a,\left|\overrightarrow{B F}_2\right|=2 a\), \(|\overrightarrow{A B}|=\dfrac{10}{3} a\),
\(\cos \angle A B F_2=\dfrac{\left|\overrightarrow{A F_2}\right|}{|\overrightarrow{A B}|}=\dfrac{2 a}{\dfrac{10}{3} a}=\dfrac{3}{5}\).
在\(△F_1 BF_2\)中, \(4 c^2=4 a^2+16 a^2-2 \times 2 a \times 4 a \times \dfrac{3}{5}\),得到 \(e^2=\dfrac{c^2}{a^2}=\dfrac{13}{5}\),
即 \(c^2=\dfrac{13}{5} a^2\),所以 \(b^2=c^2-a^2=\dfrac{13}{5} a^2-a^2=\dfrac{8}{5} a^2\).
所以 \(\dfrac{b}{a}=\sqrt{\dfrac{\dfrac{8}{5} a^2}{a^2}}=\dfrac{2 \sqrt{10}}{5}\),
故该双曲线的渐近线方程为 \(y=\pm \dfrac{2 \sqrt{10}}{5} x\).
【C组---拓展题】
1.如图,在\(△ABC\)中,已知\(∠BAC=120°\),其内切圆与\(AC\)边相切于点\(D\),延长\(BA\)到\(E\),使\(BE=BC\),连接\(CE\),设以\(E\),\(C\)为焦点且经过点\(A\)的椭圆的离心率为\(e_1\),以\(E\),\(C\)为焦点且经过点\(A\)的双曲线的离心率为\(e_2\),则当 \(\dfrac{2}{e_1}+\dfrac{1}{e_2}\)取最大值时, \(\dfrac{A D}{D C}\)的值为\(\underline{\quad \quad}\).
2.已知椭圆与双曲线有公共焦点\(F_1\),\(F_2\),\(F_1\)为左焦点,\(F_2\)为右焦点,\(P\)点为它们在第一象限的一个交点,且 \(\angle F_1 P F_2=\dfrac{\pi}{4}\),设\(e_1\),\(e_2\)分别为椭圆双曲线离心率,则 \(\dfrac{1}{e_1}+\dfrac{1}{e_2}\)的最大值为\(\underline{\quad \quad}\).
参考答案
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答案 \(\dfrac{1}{6}\)
解析 如图,设\(M\),\(G\)分别是\(BC\),\(BE\)与圆的切点.
设\(AG=AD=1\),\(CD=CM=GE=m\),\((m>1)\)
根据圆的切线性质,可得\(AC=1+m\),\(AE=GE-AG=m-1\)
在\(△ABC\)中, \(C E^2=C A^2+A E^2-2 C A \cdot E A \cos 60^{\circ}=m^2+3\)
以\(E\),\(C\)为焦点且经过点\(A\)的椭圆的离心率为 \(e_1=\dfrac{\sqrt{m^2+3}}{2 m}\);
以\(E\),\(C\)为焦点且经过点\(A\)的双曲线的离心率为 \(e_2=\dfrac{\sqrt{m^2+3}}{2}\)
则 \(\dfrac{2}{e_1}+\dfrac{1}{e_2}=\dfrac{4 m+2}{\sqrt{m^2+3}}=\sqrt{\dfrac{16 m^2+16 m+4}{m^2+3}}\)
令 \(f(m)=\dfrac{16 m^2+16 m+4}{m^2+3}\),
方法一 则 \((16-y) m^2+16 m+(4-3 y)=0(*)\)
当\(y=16\)时,方程\((*)\)有解,满足条件;
当\(y≠16\)时,若方程\((*)\)有解,
则\(△=16^2-4(16-y)(4-3y)≥0\),解得 \(y \in\left[0, \dfrac{52}{3}\right]\),
故当\(m=6\)时, \(\dfrac{2}{e_1}+\dfrac{1}{e_2}\)取最大值 \(\sqrt{\dfrac{52}{3}}\),
此时 \(\dfrac{A D}{D C}=\dfrac{1}{6}\)
方法二 \(f(m)=\dfrac{16 m^2+16 m+4}{m^2+3}=16+4 \cdot \dfrac{4 m-11}{m^2+3}\),
设\(t=4m-11\),则 \(m=\dfrac{t+11}{4}\), \(\therefore \dfrac{4 m-11}{m^2+3}=\dfrac{t}{\dfrac{(t+11)^2}{16}+3}=\dfrac{16 t}{t^2+22 t+169}=\dfrac{16}{t+\dfrac{169}{t}+22} \leq \dfrac{16}{26+22}=\dfrac{1}{3}\),
当\(t=13\),即\(m=6\)时取到等号,
\(\therefore f(m)=16+4 \cdot \dfrac{4 m-11}{m^2+3} \leq \dfrac{52}{3}\),
当\(m=6\)时, \(\dfrac{2}{e_1}+\dfrac{1}{e_2}\)取最大值 \(\sqrt{\dfrac{52}{3}}\),此时 \(\dfrac{A D}{D C}=\dfrac{1}{6}\),
故答案为: \(\dfrac{1}{6}\)
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答案 \(2 \sqrt{2}\)
解析 设椭圆与双曲线的标准方程分别为: \(\dfrac{x^2}{a_1^2}+\dfrac{y^2}{b_1^2}=1\), \(\dfrac{x^2}{a_2^2}-\dfrac{y^2}{b_2^2}=1\).
且 \(c^2=a_1^2-b_1^2=a_2^2+b_2^2\), \(a_1,a_2,b_1,b_2>0\).
设\(|PF_1 |=m\),\(|PF_2 |=n\),
则\(m+n=2a_1\),\(m-n=2a_2\).解得:\(m=a_1+a_2\),\(n=a_1-a_2\).
\((2 c)^2=m^2+n^2-2 m n \cos \dfrac{\pi}{4}\),
\(\therefore 4 c^2=\left(2 a_1\right)^2-\left(a_1+a_2\right)\left(a_1-a_2\right)(2+\sqrt{2})\),
\(\therefore 4=4 \dfrac{1}{e_1^2}-(2+\sqrt{2})\left(\dfrac{1}{e_1^2}-\dfrac{1}{e_2^2}\right)\),化为: \(\dfrac{2-\sqrt{2}}{e_1^2}+\dfrac{2+\sqrt{2}}{e_2^2}=4\).
令 \(\vec{\alpha}=\left(\dfrac{\sqrt{2-\sqrt{2}}}{e_1}, \dfrac{\sqrt{2+\sqrt{2}}}{e_2}\right)\), \(\vec{\beta}=\left(\dfrac{1}{\sqrt{2-\sqrt{2}}}, \dfrac{1}{\sqrt{2+\sqrt{2}}}\right)\),
\(\because|\vec{\alpha} \cdot \vec{\beta}| \leq|\vec{\alpha}| \cdot|\vec{\beta}|\),
\(\therefore\left(\dfrac{1}{e_1}+\dfrac{1}{e_2}\right)^2 \leq\left(\dfrac{2-\sqrt{2}}{e_1^2}+\dfrac{2+\sqrt{2}}{e_2^2}\right)\left(\dfrac{1}{2-\sqrt{2}}+\dfrac{1}{2+\sqrt{2}}\right)\).
\(\therefore \dfrac{1}{e_1}+\dfrac{1}{e_2} \leq \sqrt{4 \times \dfrac{4}{2}}=2 \sqrt{2}\).当且仅当 \(\dfrac{e_1}{e_2}=3-2 \sqrt{2}\)时取等号.