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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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选择性必修第一册同步巩固,难度2颗星!
基础知识
椭圆的几何性质
焦点的位置 | 焦点在$x$轴上 | 焦点在$y$轴上 |
图形 | ||
标准方程 | $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)$ | $\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1(a>b>0)$ |
范围 | $-a≤x≤a$且$-b≤y≤b$ | $-b≤x≤b$且$-a≤y≤a$ |
顶点 | $A_1 (-a ,0)$、$A_2 (a ,0)$、 $B_1 (0 ,-b)$、$B_2 (0 ,b)$ | $A_1 (0 ,-a)$、$A_2 (0 ,a)$、$B_1 (-b ,0)$、$B_2 (b ,0)$ |
轴长 | 短轴长$2b$,长轴长$2a$ | |
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 | $a^2=b^2+c^2$ | |
离心率 | $e=\dfrac{c}{a}=\sqrt{1-\dfrac{b^2}{a^2}}(e<1)$ |
注 离心率\(e\)越小,椭圆越偏平.
如下图, \(e=\dfrac{c}{a}=\tan \alpha\),则\(e\)越小, \(\tanα\)越小,\(α\)越小,椭圆越偏平.
【例】求椭圆 \(25x^2+16y^2=400\)的长轴长、短轴长、离心率、焦点坐标和顶点坐标.
解析 将方程变形为 \(\dfrac{y^2}{25}+\dfrac{x^2}{16}=1\),得\(a=5\),\(b=4\),所以\(c=3\).
故椭圆的长轴长和短轴长分别为\(2a=10\)和\(2b=8\),离心率 \(e=\dfrac{c}{a}=\dfrac{3}{5}\),
焦点坐标为\(F_1 (0,-3)\),\(F_2 (0,3)\),
顶点坐标为\(A_1 (0,-5)\),\(A_2 (0,5)\),\(B_1 (-4,0)\),\(B_2 (4,0)\).
一些常见结论
① 通径:过焦点且垂直于长轴的弦,其长度为 \(\dfrac{2 b^2}{a}\);
② \(P\)是椭圆上一点,当运动到短轴端点时,\(∠F_1 PF_2\)取到最大;
③ 焦点三角形面积 \(S_{\triangle P F_1 F_2}=b^2 \tan \dfrac{\angle F_1 P F_2}{2}\);
④ 焦半径\(PF_1=a+ex_0\),\(PF_2=a-ex_0\)(椭圆第二定义证明);
⑤ 椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的参数方程 \(\left\{\begin{array}{l} x=a \cdot \cos \theta \\ y=b \cdot \sin \theta \end{array}\right.\) (\(θ\)为参数).
基本方法
【题型1】椭圆的简单几何性质
【典题1】已知椭圆 \(\dfrac{x^2}{2 m^2-n}+\dfrac{y^2}{n-m^2}=1\)的焦点在\(x\)轴上,若椭圆的短轴长为\(4\),则\(n\)的取值范围是( )
A.\((12,+∞)\) \(\qquad \qquad\) B.\((4,12)\) \(\qquad \qquad\)C.\((4,6)\) \(\qquad \qquad\) D.\((6,+∞)\)
解析 依题意得 \(2m^2-n>n-m^2>0\),得 \(\dfrac{3}{2} m^2>n>m^2\),
且 \(n-m^2=4\),得 \(m^2=n-4\),
则 \(\dfrac{3}{2}(n-4)>n>n-4\),得到\(n>12\),
故选:\(A\).
【典题2】若椭圆 \(\dfrac{x^2}{9}+\dfrac{y^2}{m+4}=1\)的焦距为\(2\),则其离心率是\(\underline{\quad \quad}\) .
解析 ①当椭圆焦点在\(x\)轴上时, \(a^2=9\),\(b^2=m+4\),得 \(c=\sqrt{5-m}\),
\(∴\)焦距 \(2 c=2 \sqrt{5-m}=2\),解之得\(m=4\),
则离心率 \(e=\dfrac{c}{a}=\dfrac{1}{3}\);
②椭圆焦点在\(y\)轴上时, \(a^2=m+4\),\(b^2=9\),得 \(c=\sqrt{5-m}\),
焦距 \(2 c=2 \sqrt{m-5}=2\),解之得\(m=6\),
则离心率 \(e=\dfrac{c}{a}=\dfrac{1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}\);
综上所述,离心率是 \(\dfrac{1}{3}\)或 \(\dfrac{\sqrt{10}}{10}\).
点拨 要注意确定椭圆的焦点位置和\(a\)值.
巩固练习
1.椭圆 \(6x^2+y^2=6\)的长轴的端点坐标是( )
A.\((-1,0)\),\((1,0)\) \(\qquad \qquad \qquad\) B.\((-6,0)\),\((6,0)\)
C. \((-\sqrt{6}, 0)\), \((\sqrt{6}, 0)\) \(\qquad \qquad\) D. \((0,-\sqrt{6})\), \((0, \sqrt{6})\)
2.椭圆 \(9x^2+y^2=36\)的短轴长为\(\underline{\quad \quad}\).
3.已知焦点在\(x\)轴上的椭圆 \(\dfrac{x^2}{2}+\dfrac{y^2}{m}=1\)的离心率为\(\dfrac{1}{2}\),则\(m=\)\(\underline{\quad \quad}\) .
4.已知椭圆 \(\dfrac{x^2}{5}+\dfrac{y^2}{m}=1\)的离心率 \(e=\dfrac{\sqrt{10}}{5}\),求\(m\)的值.
参考答案
- 答案 \(D\)
- 答案 \(4\)
解析 原方程可化为 \(\dfrac{x^2}{4}+\dfrac{y^2}{36}=1\),于是\(b^2=4\),\(b=2\),从而短轴长为\(2b=4\). - 答案 \(\dfrac{3}{2}\)
解析 由题意知 \(a^2=2\),\(b^2=m\),\(∴c^2=2-m\).
\(\therefore \dfrac{\sqrt{2-m}}{\sqrt{2}}=\dfrac{1}{2}\), \(\therefore m=\dfrac{3}{2}\). - 答案 \(m=3\)或 \(m=\dfrac{25}{3}\)
解析 当焦点在\(x\)轴上时, \(a^2=5\),\(b^2=m\)
\(∴c^2=a^2-b^2=5-m\).
又 \(\because e=\dfrac{\sqrt{10}}{5}\), \(\therefore \dfrac{5-m}{5}=\left(\dfrac{\sqrt{10}}{5}\right)^2\),\(∴m=3\).
当焦点在\(y\)轴上时, \(a^2=m\),\(b^2=5\)
\(∴c^2=a^2-b^2=m-5\).
又 \(\because e=\dfrac{\sqrt{10}}{5}\), \(\therefore \dfrac{m-5}{m}=\left(\dfrac{\sqrt{10}}{5}\right)^2\), \(\therefore m=\dfrac{25}{3}\).
故\(m=3\)或 \(m=\dfrac{25}{3}\).
【题型2】利用椭圆的几何性质求椭圆方程
【典题1】 求满足下列条件的椭圆的标准方程:
(1)与椭圆 \(4x^2+9y^2=36\)有相同的焦点,且离心率为 \(\dfrac{\sqrt{5}}{5}\);
(2)已知椭圆的对称轴是坐标轴,\(O\)为坐标原点,\(F\)是一个焦点,\(A\)是一个顶点,椭圆的长轴长是\(6\),且 \(\cos \angle O F A=\dfrac{2}{3}\).
解析 (1)把方程 \(4x^2+9y^2=36\)化成 \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\),
则焦点在\(x\)轴,焦距为 \(2 \sqrt{5}\),则 \(c=\sqrt{5}\),
又 \(\dfrac{c}{a}=\dfrac{\sqrt{5}}{5}\), \(∴a=5\), \(b^2=a^2-c^2=5^2-5=20\).
\(∴\)所求椭圆的 方程为 \(\dfrac{x^2}{25}+\dfrac{y^2}{20}=1\).
(2)\(∵\)椭圆的长轴长是\(6\), \(\cos \angle O F A=\dfrac{2}{3}\),
\(∴\)点\(A\)不是长轴的端点(是短轴的端点).
\(∴|OF|=c\),\(|AF|=a=3\).
\(\therefore \dfrac{c}{3}=\dfrac{2}{3}\),\(∴c=2\), \(b^2=3^2-2^2=5\).
\(∴\)椭圆的方程是 \(\dfrac{x^2}{9}+\dfrac{y^2}{5}=1\)或 \(\dfrac{x^2}{5}+\dfrac{y^2}{9}=1\).
点拨 掌握椭圆中 \(a^2=b^2+c^2\),注意结合椭圆图像及其性质确定椭圆方程.
巩固练习
1.中心在原点,焦点在\(x\)轴上,若长轴长为\(18\),且两个焦点恰好将长轴三等分,则此椭圆的方程是( )
A. \(\dfrac{x^2}{81}+\dfrac{y^2}{72}=1\) \(\qquad \qquad\) B. \(\dfrac{x^2}{81}+\dfrac{y^2}{9}=1\) \(\qquad \qquad\) C. \(\dfrac{x^2}{81}+\dfrac{y^2}{45}=1\) \(\qquad \qquad\) D. \(\dfrac{x^2}{81}+\dfrac{y^2}{36}=1\)
2.求适合下列条件的椭圆的标准方程:
(1) 焦点在\(x\)轴上,椭圆过\(P(3,0)\),且 \(e=\dfrac{\sqrt{6}}{3}\);
(2) 椭圆的长轴长是短轴长的\(2\)倍,且过\(P(2,-6)\).
参考答案
- 答案 \(A\)
解析 由已知得\(2a=18\),\(2c=6\),\(∴a=9\),\(c=3\).从而 \(b^2=a^2-c^2=72\),
又焦点在\(x\)轴上,\(∴\)所求椭圆的方程为 \(\dfrac{x^2}{81}+\dfrac{y^2}{72}=1\). - 答案 (1) \(\dfrac{x^2}{9}+\dfrac{y^2}{3}=1\) (2) \(\dfrac{x^2}{148}+\dfrac{y^2}{37}=1\)或 \(\dfrac{y^2}{52}+\dfrac{x^2}{13}=1\)
解析 (1) \(∵\)过\(P(3,0)\),\(∴a=3\),
又 \(\dfrac{c}{a}=\dfrac{\sqrt{6}}{3}\), \(\therefore c=\sqrt{6}\).
\(∴b^2=a^2-c^2=3\).
此时椭圆的标准方程为 \(\dfrac{x^2}{9}+\dfrac{y^2}{3}=1\);
(2) 设椭圆标准方程为 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)或 \(\dfrac{y^2}{a^2}+\dfrac{x^2}{b^2}=1(a>b>0)\).
由已知得\(a=2b\).①
\(∵\)椭圆过\(P(2,-6)\), \(\therefore \dfrac{4}{a^2}+\dfrac{36}{b^2}=1\)或 \(\dfrac{36}{a^2}+\dfrac{4}{b^2}=1\).②
由①②得 \(a^2=148, b^2=37\)或 \(a^2=52, b^2=13\).
故所求椭圆标准方程为 \(\dfrac{x^2}{148}+\dfrac{y^2}{37}=1\)或 \(\dfrac{y^2}{52}+\dfrac{x^2}{13}=1\).
【题型3】椭圆的几何性质综合问题
【典题1】已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左右焦点分别为\(F_1\),\(F_2\),\(P\),\(Q\)为过\(F_2\)的直线与椭圆\(C\)的交点,且\(△F_1 PQ\)为正三角形,则该椭圆的离心率为\(\underline{\quad \quad}\) .
解析 \(∵△F_1 PQ\)为正三角形,\(∴PF_1=F_1 Q\),
由对称性可知\(P\),\(Q\)关于\(x\)轴对称,
\(∴\)直线\(PQ\)方程为\(x=c\),故 \(P F_2=\dfrac{b^2}{a}\),
由等边三角形性质可知 \(F_1 F_2=\sqrt{3} P F_2\),
\(\therefore 2 c=\sqrt{3} \cdot \dfrac{b^2}{a}\),即 \(2 a c=\sqrt{3} a^2-\sqrt{3} c^2\),
\(\therefore \sqrt{3} e^2+2 e-\sqrt{3}=0\),解得 \(e=\dfrac{\sqrt{3}}{3}\)或 \(e=-\sqrt{3}\)(舍).
故答案为: \(\dfrac{\sqrt{3}}{3}\)
点拨 求离心率 \(e=\dfrac{c}{a}\),由于有 \(a^2=b^2+c^2\),则只需要得到\(a,b,c\)中任意两个参数间的齐次方程便可,如本题中的 \(2 a c=\sqrt{3} a^2-\sqrt{3} c^2\)或 \(a^2=3 b^2-2 b a\)或\(a=2b\)等.
【典题2】已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左、右焦点分别为\(F_1\),\(F_2\),\(B\),\(C\)分别为椭圆的上、下顶点,直线\(BF_2\)与椭圆的另一个交点为\(D\),若 \(\cos \angle F_1 B F_2=\dfrac{7}{25}\),则直线\(CD\)的斜率为( )
A. \(\dfrac{24}{25}\) \(\qquad \qquad\) B. \(\dfrac{14}{25}\) \(\qquad \qquad\) C. \(\dfrac{12}{25}\) \(\qquad \qquad\) D. \(\dfrac{7}{25}\)
解析 在三角形\(∆F_1 BF_2\)中,
由余弦定理可得 \(\cos \angle F_1 B F_2=\dfrac{7}{25} \Rightarrow \dfrac{a^2+a^2-4 c^2}{2 a^2}=\dfrac{7}{25}\), 解得 \(\dfrac{c}{a}=\dfrac{3}{5}\),
可设\(a=5t\),\(c=3t\),则 \(b=\sqrt{a^2-c^2}=4 t\),
设\(D(m,n)\),即有 \(\dfrac{m^2}{a^2}+\dfrac{n^2}{b^2}=1\),
\(∵B(0,b)\),\(C(0,-b)\),
\(\therefore k_{B D} \cdot k_{C D}=\dfrac{n-b}{m} \cdot \dfrac{n+b}{m}=\dfrac{n^2-b^2}{m^2}=\dfrac{n^2-b^2}{a^2\left(1-\dfrac{n^2}{b^2}\right)}=-\dfrac{b^2}{a^2}=-\dfrac{16}{25}\),
由 \(k_{B D}=k_{B F_2}=-\dfrac{b}{c}=-\dfrac{4}{3}\), \(\therefore k_{C D}=\dfrac{12}{25}\).
故选:\(C\).
点拨 对斜率\(k\)的处理可考虑使用斜率公式 \(k=\dfrac{y_2-y_1}{x_2-x_1}\).
【典题3】设椭圆 \(C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的右焦点为\(F\),椭圆\(C\)上的两点\(A\)、\(B\)关于原点对称,且满足 \(\overrightarrow{F A} \cdot \overrightarrow{F B}=0\),\(|FB|≤|FA|≤2|FB|\),则椭圆\(C\)的离心率的取值范围是\(\underline{\quad \quad}\).
解析 作出椭圆的左焦点\(F’\),由椭圆的对称性可知,四边形\(AFBF’\)为平行四边形,
又 \(\overrightarrow{F A} \cdot \overrightarrow{F B}=0\),即\(FA⊥FB\),故平行四边形\(AFBF’\)为矩形,
\(∴|AB|=|FF’|=2c\),
设\(AF’=n\),\(AF=m\),
则在直角三角形\(ABF\)中,\(m+n=2a\), \(m^2+n^2=4c^2\),①
得\(mn=2b^2\),②
①÷②得 \(\dfrac{m}{n}+\dfrac{n}{m}=\dfrac{2 c^2}{b^2}\) ,令 \(\dfrac{m}{n}=t\),得 \(t+\dfrac{1}{t}=\dfrac{2 c^2}{b^2}\) ,
又由\(|FB|≤|FA|≤2|FB|\),得 \(\dfrac{m}{n}=t \in[1,2]\),
\(\therefore t+\dfrac{1}{t}=\dfrac{2 c^2}{b^2} \in\left[2, \dfrac{5}{2}\right]\),即 \(\dfrac{c^2}{b^2} \in\left[1, \dfrac{5}{4}\right]\)
即 \(\text { 即 } 1 \leq \dfrac{c^2}{b^2} \leq \dfrac{5}{4}\),得\(\dfrac{4}{5} \leq \dfrac{b^2}{c^2} \leq 1\),即 \(\dfrac{4}{5} \leq \dfrac{a^2-c^2}{c^2} \leq 1\),即 \(\dfrac{4}{5} \leq \dfrac{a^2}{c^2}-1 \leq 1\),
则 \(\dfrac{9}{5} \leq \dfrac{a^2}{c^2} \leq 2\),即 \(\dfrac{1}{2} \leq \dfrac{c^2}{a^2} \leq \dfrac{5}{9}\),得 \(\dfrac{\sqrt{2}}{2} \leq e \leq \dfrac{\sqrt{5}}{3}\),
则椭圆的离心率的取值范围是 \(\left[\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{5}}{3}\right]\).
点拨
1.求离心率 \(e=\dfrac{c}{a}\)的取值范围与求离心率\(e\)值思路相似,由于有 \(a^2=b^2+c^2\),则只需要得到a,b,c中任意两个参数间的齐次不等式便可,如本题中的 \(1 \leq \dfrac{c^2}{b^2} \leq \dfrac{5}{4}\)或 \(a^2>3 b^2-2 b a\)或\(a>2b\)等;
2.对椭圆几何性质的求解,注意对图形的分析,根据已知条件确定图形的特征.本题也可设点\(A(m,n)\)从代数的角度求解,但较于几何法略显复杂.
巩固练习
1.(多选)如图,两个椭圆 \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\), \(\dfrac{y^2}{25}+\dfrac{x^2}{9}=1\)内部重叠区域的边界记为曲线\(C\),\(P\)是曲线\(C\)上的任意一点,下列四个判断中正确命题为( )
A.\(P\)到\(F_1 (-4,0)\)、\(F_2 (4,0)\)、\(E_1 (0,-4)\)、\(E_2 (0,4)\)四点的距离之和为定值
B.曲线\(C\)关于直线\(y=x\)、\(y=-x\)均对称
C.曲线\(C\)所围区域面积必小于\(36\)
D.曲线\(C\)总长度不大于\(6π\)
2.已知点\(A(a,0)\)、\(B(0,b)\),椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)经过点 \(D(-2,-\sqrt{3})\),点\(F\)为椭圆的右焦点,若\(△FAB\)的一个内角为\(120°\),则椭圆\(C\)的方程是\(\underline{\quad \quad}\).
3.已知\(F_1\),\(F_2\)分别是椭圆 \(E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左、右焦点,\(P\)为直线\(x=a\)上一点,若\(△F_1 PF_2\)是底角为\(30°\)的等腰三角形,则椭圆\(E\)的离心率为\(\underline{\quad \quad}\) .
4.已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左、右焦点为\(F_1\) 、\(F_2\),\(O\)为坐标原点,\(M\)为椭圆上一点.\(F_1 M\)与\(y\)轴交于一点\(N\), \(|O M|=\left|O F_2\right|=\sqrt{3}|O N|\),则椭圆\(C\)的离心率为\(\underline{\quad \quad}\).
5.已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的短轴长为\(2\),上顶点为\(A\),左顶点为\(B\),左右焦点分别是\(F_1\),\(F_2\),且\(△F_1 AB\)的面积为 \(\dfrac{2-\sqrt{3}}{2}\),则椭圆的方程为\(\underline{\quad \quad}\);若点\(P\)为椭圆上的任意一点,则 \(\dfrac{1}{\left|P F_1\right|}+\dfrac{1}{\left|P F_2\right|}\)的取值范围是\(\underline{\quad \quad}\).
参考答案
-
答案 \(BC\)
解析 逐一考查所给的说法:
考虑点\(P\)不是交点的情况,若点\(P\)在椭圆 \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\)上,\(P\)到\(F_1 (-4,0)\)、\(F_2 (4,0)\)两点的距离之和为定值、到\(E_1 (0,-4)\)、\(E_2 (0,4)\)两点的距离之和不为定值,故\(A\)错;
两个椭圆关于直线\(y=x\)、\(y=-x\)均对称,曲线\(C\)关于直线\(y=x\)、\(y=-x\)均对称,
故\(B\)正确;
曲线\(C\)所围区域在边长为\(6\)的正方形内部,所以面积必小于\(36\),故\(C\)正确;
曲线\(C\)所围区域在半径为\(3\)的圆外部,所以曲线的总长度大于圆的周长:\(6π\),故\(D\)错误;
故选:\(BC\). -
答案 \(\dfrac{x^2}{8}+\dfrac{y^2}{6}=1\)
解析 如图,
由题意, \(\dfrac{4}{a^2}+\dfrac{3}{b^2}=1\),①
\(A B^2=F A^2+F B^2-2 F A \cdot F B \cdot \cos 120^{\circ}\),
即 \(a^2+b^2=(a-c)^2+a^2+a(a-c)\),②
又 \(a^2=b^2+c^2\),③
联立①②③,解得 \(a^2=8\),\(b^2=6\).
\(∴\)椭圆\(C\)的方程是 \(\dfrac{x^2}{8}+\dfrac{y^2}{6}=1\). -
答案 \(\dfrac{1}{2}\)
解析 设\(x=a\)交\(x\)轴于点\(M\),
\(∵△F_2 PF_1\)是底角为\(30°\)的等腰三角形
\(∴∠PF_2 F_1=120^∘\),\(|PF_2 |=|F_2 F_1 |=2c\),\(∠PF_2 M=60^∘\),可得\(|F_2 M|=c=a-c\),
∴椭圆E的离心率为 \(e=\dfrac{c}{a}=\dfrac{1}{2}\),
故答案为: \(\dfrac{1}{2}\).
-
答案 \(\sqrt{3}-1\)
解析 因为 \(|O M|=\left|O F_2\right|=\sqrt{3}|O N|\),所以\(∠F_1 MF_2=90°\),
设\(MF_1=m\),\(MF_2=n\).
如图所示,由题意:\(Rt△MF_1 F_2∽Rt△ONF_2\),\(|OF_2 |=\sqrt{3}|ON|\),
可得 \(\dfrac{M F_1}{M F_2}=\dfrac{O N}{O F_2}=\dfrac{m}{n}=\dfrac{\sqrt{3}}{3}\),则 \(m+n=2 a\), \(m^2+n^2=4 c^2\), \(n=\sqrt{3} m\).
可得 \(m=(\sqrt{3}-1) a\), \(n=(3-\sqrt{3}) a\),
\(\therefore(\sqrt{3}-1)^2 a^2+(3-\sqrt{3})^2 a^2=4 c^2\).
\(\therefore(4-2 \sqrt{3}) a^2=c^2\),化为: \(\dfrac{c}{a}=\sqrt{3}-1\). -
答案 \(\dfrac{x^2}{4}+y^2=1\);\([1,4]\)
解析 由已知可得\(2b=2\),即\(b=1\),
\(∵△F_1 AB\)的面积为 \(\dfrac{2-\sqrt{3}}{2}\),
\(\therefore \dfrac{1}{2}(a-c) b=\dfrac{2-\sqrt{3}}{2}\),得 \(a-c=2-\sqrt{3}\);
\(∵a^2-c^2=b^2=1\);\(∴a=2\),\(c=\sqrt{3}\) .
可得椭圆方程为 \(\dfrac{x^2}{4}+y^2=1\);
\(\therefore \dfrac{1}{\left|P F_1\right|}+\dfrac{1}{\left|P F_2\right|}=\dfrac{\left|P F_1\right|+\left|P F_2\right|}{\left|P F_1\right| \cdot\left|P F_2\right|}=\dfrac{2 a}{\left|P F_1\right|\left(2 a-\left|P F_1\right|\right)}\).
令\(|PF_1 |=m\),则 \(2-\sqrt{3} \leq m \leq 2+\sqrt{3}\).
\(\therefore \dfrac{1}{\left|P F_1\right|}+\dfrac{1}{\left|P F_2\right|}=\dfrac{4}{m(4-m)}=\dfrac{4}{-m^2+4 m}\),
\(\because 2-\sqrt{3} \leq m \leq 2+\sqrt{3}\), \(\therefore 1 \leq-m^2+4 m \leq 4\);
\(\therefore 1 \leq \dfrac{1}{\left|P F_1\right|+\left|P F_2\right|} \leq 4\).
故答案为:\(\dfrac{x^2}{4}+y^2=1\);\([1,4]\).
分层练习
【A组—基础题】
1.已知椭圆\(C\)的标准方程为 \(\dfrac{x^2}{16}+\dfrac{y^2}{25}=1\),下列说法正确的是( )
A.椭圆\(C\)的焦点在\(x\)轴上\(\qquad \qquad\) B.椭圆\(C\)的焦距为\(3\)
C.椭圆\(C\)的离心率为 \(\dfrac{3}{5}\) \(\qquad \qquad\) D.椭圆\(C\)的右顶点坐标为\((5,0)\)
2.曲线 \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\)与曲线 \(\dfrac{x^2}{9-k}+\dfrac{y^2}{25-k}=1(0<k<9)\)的关系是( ).
A.有相等的焦距,相同的焦点 \(\qquad \qquad\) B.有相等的焦距,不同的焦点
C.有不等的焦距,不同的焦点 \(\qquad \qquad\) D.以上都不对
3.若焦点在\(x\)轴上的椭圆 \(\dfrac{x^2}{2}+\dfrac{y^2}{n}=1\)的离心率为 \(\dfrac{2}{3}\),则\(n=\)( )
A. \(\dfrac{18}{5}\) \(\qquad \qquad\) B. \(\dfrac{8}{9}\) \(\qquad \qquad\) C. \(\dfrac{10}{9}\) \(\qquad \qquad\) D. \(\dfrac{8}{5}\)
4.在平面直角坐标系\(xOy\)中,已知椭圆 \(E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的右焦点为\(F(c,0)\),若\(F\)到直线\(2bx-ay=0\)的距离为 \(\dfrac{\sqrt{2}}{2} c\),则\(E\)的离心率为( )
A. \(\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad\) B. \(\dfrac{1}{2}\) \(\qquad \qquad\)C. \(\dfrac{\sqrt{2}}{2}\) \(\qquad \qquad\) D. \(\dfrac{\sqrt{2}}{3}\)
5.直线 \(x-\sqrt{3} y+\sqrt{3}=0\)经过椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左焦点\(F\),交椭圆于\(A\),\(B\)两点,交\(y\)轴于\(C\)点,若 \(\overrightarrow{F A}=3 \overrightarrow{C A}\),则该椭圆的离心率是( )
A. \(\dfrac{\sqrt{3}-1}{2}\) \(\qquad \qquad\) B. \(\sqrt{3}-1\) \(\qquad \qquad\) C. \(2 \sqrt{2}-2\) \(\qquad \qquad\) D. \(\sqrt{2}-1\)
6.(多选)如图,椭圆Ⅰ与Ⅱ有公共的左顶点和左焦点,且椭圆Ⅱ的右顶点为椭圆Ⅰ的中心.设椭圆Ⅰ与Ⅱ的长半轴长分别为\(a_1\)和\(a_2\),半焦距分别为\(c_1\)和\(c_2\),离心率分别为\(e_1\),\(e_2\),则下列结论正确的是( )
A.\(a_1+c_1>2(a_2+c_2)\) \(\qquad \qquad\) B.\(a_1-c_1=a_2-c_2\)
C.\(a_1 c_2>a_2 c_1\) \(\qquad \qquad \qquad\) D. \(e_1=\dfrac{e_2+1}{2}\) \(\qquad \qquad\) E.椭圆Ⅱ比椭圆Ⅰ更扁
7.(多选)已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的右焦点为\(F\),点\(P\)在椭圆\(C\)上,点\(Q\)在圆 \(E:(x+3)^2+(y-4)^2=4\)上,且圆\(E\)上的所有点均在椭圆\(C\)外,若\(|PQ|-|PF|\)的最小值为\(2 \sqrt{5}-6\),且椭圆\(C\)的长轴长恰与圆\(E\)的直径长相等,则下列说法正确的是( )
A.椭圆\(C\)的焦距为\(2\) \(\qquad \qquad \qquad \qquad\) B.椭圆\(C\)的短轴长为 \(\sqrt{3}\) \(\qquad \qquad\)
C.\(|PQ|+|PF|\)的最小值为 \(2 \sqrt{5}\) \(\qquad \qquad\) D.过点\(F\)的圆\(E\)的切线斜率为 \(\dfrac{-4 \pm \sqrt{7}}{3}\)
8.若椭圆的短轴长为 \(4 \sqrt{5}\),它的一个焦点是 \((0,2 \sqrt{15})\),则该椭圆的标准方程为\(\underline{\quad \quad}\).
9.已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左、右焦点分别为\(F_1\),\(F_2\),点\(P\)为椭圆上不同于左、右顶点的任意一点,\(I\)为\(△PF_1 F_2\)的内心,且 \(S_{\triangle I P F_1}=\lambda S_{\Delta I F_1 F_2}-S_{\triangle I P F_2}\),若椭圆的离心率为\(e\),则\(λ=\)\(\underline{\quad \quad}\) .
10.已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左、右焦点为\(F_1\),\(F_2\),上顶点为\(A\),点\(P\)为第一象限内椭圆上的一点, \(\left|P F_1\right|+\left|P F_2\right|=4\left|F_1 F_2\right|\), \(S_{\triangle P F_1 A}=2 S_{\triangle P F_1 F_2}\),则直线\(PF_1\)的斜率为\(\underline{\quad \quad}\).
11.已知椭圆 \(E: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左焦点为\(F\),\(A\)、\(B\)两点是椭圆\(E\)上关于\(y\)轴对称的点,若\(△ABF\)能构成一个内角为 \(\dfrac{2 \pi}{3}\)的等腰三角形,则椭圆\(E\)的离心率\(e=\)\(\underline{\quad \quad}\).
12.若椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=11\)的焦点在\(x\)轴上,过点 \(\left(1, \dfrac{1}{2}\right)\)作圆 \(x^2+y^2=1\)的切线,切点分别为\(A\)、\(B\),直线\(AB\)恰好经过椭圆的右焦点和上顶点,则椭圆方程是\(\underline{\quad \quad}\) .
参考答案
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答案 \(C\)
解析 \(∵25>16\),故椭圆焦点在\(y\)轴上,故\(A\)错误;
\(∵a^2=25\),\(b^2=16\),\(∴c^2=9\),\(c=3\),故椭圆焦距为\(2c=6\),故\(B\)错误;
椭圆离心率 \(e=\dfrac{c}{a}=\dfrac{3}{5}\),故\(C\)正确;
椭圆右顶点为\((4,0)\),故\(D\)错误.
故选:\(C\). -
答案 \(B\)
解析 \(∵0<k<9\),\(∴\)曲线 \(\dfrac{x^2}{9-k}+\dfrac{y^2}{25-k}=1(0<k<9)\)是焦点在\(y\)轴上的椭圆.
\(∴c^2=(25-k)-(9-k)=16\).
又 \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\)是焦点在\(x\)轴上的椭圆且\(c^2=25-9=16\),
因 此两曲线有相等的焦距,但焦点不相同. -
答案 \(C\)
解析 \(∵\)椭圆焦点在\(x\)轴上,\(∴0<n<2\),
又离心率 $e=\dfrac{\sqrt{2-n}}{\sqrt{2}}=\dfrac{2}{3} $,故 \(n=\dfrac{10}{9}\),
故选:\(C\). -
答案 \(A\)
解析 由\(F\)到直线\(2bx-ay=0\)的距离为 \(\dfrac{\sqrt{2}}{2} c\), \(\dfrac{|2 b c|}{\sqrt{4 b^2+a^2}}=\dfrac{\sqrt{2}}{2}c\),可得\(a=2b\).
即 \(4(a^2-c^2)=a^2\),解得 \(e=\dfrac{\sqrt{3}}{2}\),
故选:\(A\). -
答案 \(B\)
解析 由直线方程可的\(F(-\sqrt{3},0)\),\(C(0,1)\),则\(c=\sqrt{3}\),
又因为 \(\overrightarrow{F A}=3 \overrightarrow{C A}\),即\(|FA|=3|CA|\),
过\(A\)作\(x\)轴垂线,根据相似三角形可得 \(A\left(\dfrac{\sqrt{3}}{2}, \dfrac{3}{2}\right)\),
则 \(2 a=\sqrt{\left(\dfrac{\sqrt{3}}{2}+\sqrt{3}\right)^2+\left(\dfrac{3}{2}\right)^2}+\sqrt{\left(\dfrac{\sqrt{3}}{2}-\sqrt{3}\right)^2+\left(\dfrac{3}{2}\right)^2}=3+\sqrt{3}\),
所以 \(e=\dfrac{2 c}{2 a}=\dfrac{2 \sqrt{3}}{3+\sqrt{3}}=\sqrt{3}-1\),
故选:\(B\).
-
答案 \(ABD\)
解析 由题图知,\(a_1=2a_2\),\(c_1>2c_2\),
\(∴a_1+c_1>2(a_2+c_2)\),正确;故\(A\)正确;
\(∵\)椭圆Ⅰ与Ⅱ有公共的左顶点和左焦点,且椭圆Ⅱ的右顶点为椭圆Ⅰ的中心,
\(∴a_1-c_1=a_2-c_2\)正确;故\(B\)正确;
\(a_1 c_2<a_2 c_1\)正确;
\(a_1 c_2>a_2 c_1\)不正确.故\(C\)不正确;
由图知,\(c_1=a_2+c_2\);
\(\therefore e_1=\dfrac{c_1}{a_1}=\dfrac{a_2+c_2}{2 a_2}\); \(\dfrac{e_2+1}{2}=\dfrac{\dfrac{c_2}{a_2+1}}{2}=\dfrac{a_2+c_2}{2 a_2}\);
\(\therefore e_1=\dfrac{e_2+1}{2}\)正确;故\(D\)正确;
\(e_1=\dfrac{c_1}{a_1}>\dfrac{2 c_2}{2 a_2}=e_2\);\(∴\)椭圆Ⅰ比椭圆Ⅱ更扁;故\(E\)不正确;
∴正确的为\(ABD\).
故选:\(ABD\). -
答案 \(AD\)
解析 根据题意,作出如下所示的图形,
\(∵\)椭圆\(C\)的长轴长与圆\(E\)的直径长相等,\(∴2a=4\),\(a=2\),
设椭圆的左焦点为\(F'(-c,0)\),由椭圆的定义可知,\(|PF’|+|PF|=2a=4\),
\(\therefore|P Q|-|P F|=|P Q|-\left(4-\left|P F^{\prime}\right|\right) \geq\left|Q F^{\prime}\right|-4 \geq\left|E F^{\prime}\right|-2-4=2 \sqrt{5}-6\),
\(\therefore\left|E F^{\prime}\right|=2 \sqrt{5}=\sqrt{(-3+c)^2+(4-0)^2}\),解得\(c=1\)或\(5\)(\(∵c<a\),\(∴\)舍\(5\)),
\(∴\)椭圆\(C\)的焦距为\(2\),即\(A\)正确;
由 \(b=\sqrt{a^2-c^2}=\sqrt{4-1}=\sqrt{3}\),得椭圆\(C\)的短轴长为\(2\sqrt{3}\),即\(B\)错误;
\(|P Q|+|P F| \geq|E F|=\sqrt{(-3-1)^2+(4-0)^2}=4 \sqrt{2}\),即\(C\)错误;
设过点\(F\)的圆\(E\)的切线方程为\(y=k(x-1)\),
则 \(\dfrac{|k(-3-1)-4|}{\sqrt{k^2+1}}=2\),解得 \(k=\dfrac{-4 \pm \sqrt{7}}{3}\),即\(D\)正确.
故选:\(AD\). -
答案 \(\dfrac{y^2}{80}+\dfrac{x^2}{20}=1\)
解析 依题意椭圆的焦点在\(y\)轴上,且 \(c=2 \sqrt{15}\),
又知 \(2 b=4 \sqrt{5}\),所以 \(b=2 \sqrt{5}\),于是 \(a=\sqrt{b^2+c^2}=4 \sqrt{5}\),
所以椭圆的标准方程是 \(\dfrac{y^2}{80}+\dfrac{x^2}{20}=1\). -
答案 \(\dfrac{1}{e}\)
解析 设\(△PF_1 F_2\)的内切圆半径为\(r\),
则 \(S_{\triangle I P F_1}=\dfrac{1}{2}\left|P F_1\right| \cdot r\),\(S_{\triangle I P F_2}=\dfrac{1}{2}\left|P F_2\right| \cdot r\),\(S_{\Delta I F_1 F_2}=\dfrac{1}{2}\left|F_1 F_2\right| \cdot r\),
\(\because S_{\triangle I P F_1}+S_{\triangle I P F_2}=\lambda S_{\Delta I F_1 F_2}\),
\(\therefore \dfrac{1}{2}\left|P F_1\right| \cdot r+\dfrac{1}{2}\left|P F_2\right| \cdot r=\lambda \cdot \dfrac{1}{2}\left|F_1 F_2\right| \cdot r\),
可得\(|PF_1 |+|PF_2 |=λ|F_1 F_2 |\).
\(∴2a=λ2c\),解得: \(\lambda=\dfrac{1}{e}\). -
答案 \(\dfrac{\sqrt{15}}{5}\)
解析 \(∵|PF_1 |+|PF_2 |=4|F_1 F_2 |\),\(∴2a=4×2c\),即\(a=4c\), \(\therefore b=\sqrt{a^2-c^2}=\sqrt{15} c\),
由题可知,点\(A(0,b)\),\(F_1 (-c,0)\),\(F_2 (c,0)\),
设直线\(PF_1\)的方程为\(y=k(x+c)\),
\(\because S_{\triangle P F_1 A}=2 S_{\triangle P F_1 F_2}\), \(\therefore d_{A-P F_1}=2 d_{F_2-P F_1}\),
由点到直线的距离公式有, \(d_{A-P F_1}=\dfrac{|k c-b|}{\sqrt{k^2+1}}\), \(d_{F_2-P F_1}=\dfrac{|2 k c|}{\sqrt{k^2+1}}\),
\(∴|kc-b|=2|2kc|\),即 \(|k c-\sqrt{15} c|=4|k c|\),
\(∵c≠0\), \(\therefore|k-\sqrt{15}|=4|k|\),解得 \(k=\dfrac{\sqrt{15}}{5}\)或 \(-\dfrac{\sqrt{15}}{3}\)(舍负).
故答案为: \(\dfrac{\sqrt{15}}{3}\). -
答案 \(\sqrt{3}-1\)
解析 如图,设椭圆\(E\)的右焦点为\(F’\),连接\(BF’\),
则四边形\(FABF’\)为等腰梯形,其中 \(\angle F A B=\dfrac{2 \pi}{3}\),
\(\therefore \angle F^{\prime} F B=\dfrac{\pi}{6}\), \(\angle F F^{\prime} B=\dfrac{\pi}{3}\), \(\angle F B F^{\prime}=\dfrac{\pi}{2}\),
\(∴\)在焦点三角形\(△FF’B\)中, \(e=\dfrac{F F^{\prime}}{B F+B F^{\prime}}=\dfrac{\sin \dfrac{\pi}{2}}{\sin \dfrac{\pi}{6}+\sin \dfrac{\pi}{2}}=\dfrac{2}{\sqrt{3}+1}=\sqrt{3}-1\),
即椭圆\(E\)的离心率为 \(\sqrt{3}-1\). -
答案 \(\dfrac{x^2}{5}+\dfrac{y^2}{4}=1\)
解析 设过点 \(\left(1, \dfrac{1}{2}\right)\)作圆 \(x^2+y^2=1\)的切线为 \(l: y-\dfrac{1}{2}=k(x-1)\),即 \(k x-y-k+\dfrac{1}{2}=0\)
①当直线\(l\)与\(x\)轴垂直时,\(k\)不存在,直线方程为\(x=1\),
恰好与圆 \(x^2+y^2=1\)相切于点\(A(1,0)\);
②当直线\(l\)与\(x\)轴不垂直时,
原点到直线\(l\)的距离为: \(d=\dfrac{\left|-k+\dfrac{1}{2}\right|}{\sqrt{k^2+1}}=1\),解之得 \(k=-\dfrac{3}{4}\),
此时直线\(l\)的方程为 \(y=-\dfrac{3}{4} x+\dfrac{5}{4}\),\(l\)切圆 \(x^2+y^2=1\)相切于点 \(B\left(\dfrac{3}{5}, \dfrac{4}{5}\right)\);
因此,直线\(AB\)斜率为 \(k_1=\dfrac{0-\dfrac{4}{5}}{1-\dfrac{3}{5}}=-2\),直线\(AB\)方程为\(y=-2(x-1)\)
\(∴\)直线\(AB\)交\(x\)轴交于点\(A(1,0)\),交\(y\)轴于点\(C(0,2)\).
椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)的右焦点为\((0,1)\),上顶点为\((0,2)\)
\(∴c=1\),\(b=2\),可得 \(a^2=b^2+c^2=5\),
故椭圆方程为 \(\dfrac{x^2}{5}+\dfrac{y^2}{4}=1\).
【B组—提高题】
1.已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左右焦点分别为\(F_1\),\(F_2\),点\(A\)是椭圆上一点,线段\(AF_1\)的垂直平分线与椭圆的一个交点为\(B\),若 \(\overrightarrow{A B}=3 \overrightarrow{F_2 B}\),则椭圆\(C\)的离心率为\(\underline{\quad \quad}\) .
2.已知平面内与两定点距离的比为常数\(k\)(\(k>0\),且\(k≠1\))的点的轨迹是圆,这个圆称为阿波罗尼斯圆.现有椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\),\(A\),\(B\)为长轴端点,\(C\),\(D\)为短轴端点,动点\(M\)满足 \(\dfrac{|M A|}{|M B|}=2\),\(△MAB\)面积的最大值为\(8\),\(△MCD\)面积的最小值为\(1\),则椭圆的离心率为\(\underline{\quad \quad}\) .
参考答案
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答案 \(\dfrac{\sqrt{3}}{3}\)
解析 如图所示,线段\(AF_1\)的垂直平分线与椭圆的一个交点为\(B\),连接\(BF_1\).
则\(|AB|=|BF_1 |\).
\(∵\overrightarrow{A B}=3 \overrightarrow{F_2 B}\),\(|BF_1 |+|BF_2 |=2a\), \(\therefore\left|B F_2\right|=\dfrac{a}{2}\),\(|AF_2 |=a\).
\(∴\)点\(A\)是椭圆短轴的一个端点,不妨设为上端点.
作\(BC⊥x\)轴,垂足为点\(C\).则\(△AOF_2∽△BCF_2\).
\(\therefore \dfrac{|B C|}{|A O|}=\dfrac{\left|C F_2\right|}{\left|O F_2\right|}=\dfrac{\left|B F_2\right|}{\left|A F_2\right|}=\dfrac{1}{2}\).
\(\therefore y_B=-\dfrac{1}{2} b\), \(\left|C F_2\right|=\dfrac{1}{2} c\). \(\therefore B\left(\dfrac{3 c}{2},-\dfrac{b}{2}\right)\).
代入椭圆方程可得: \(\dfrac{9 c^2}{4 a^2}+\dfrac{1}{4}=1\),解得 \(\dfrac{c^2}{a^2}=\dfrac{1}{3}\).
\(\therefore e=\dfrac{c}{a}=\dfrac{\sqrt{3}}{3}\). -
答案 \(\dfrac{\sqrt{3}}{2}\)
解析 由椭圆的方程可得\(A(-a,0)\),\(B(a,0)\),\(C(0,b)\),\(D(0,-b)\),设\(M(x,y)\),
则点\(M\)满足 \(\dfrac{|M A|}{|M B|}=2\),
所以可得 \(\dfrac{\sqrt{(x+a)^2+y^2}}{\sqrt{(x-a)^2+y^2}}=2\),整理可得 \(x^2+y^2-\dfrac{10}{3} a x+a^2=0\),
圆心坐标为 \(\left(\dfrac{5}{3} a, 0\right)\),半径 \(r=\dfrac{4}{3} a\),
因为\(△MAB\)面积的最大值为\(8\),\(△MCD\)面积的最小值为\(1\),
所以 \(\left\{\begin{array}{l} \dfrac{1}{2} \cdot 2 a \cdot \dfrac{4}{3} a=8 \\ \dfrac{1}{2} \cdot 2 b \cdot\left(\dfrac{5}{3} a-\dfrac{4}{3} a\right)=1 \end{array}\right.\)解得: \(a^2=6, b^2=\dfrac{3}{2}\),
所以椭圆的离心率 \(e=\dfrac{c}{a}=\sqrt{1-\dfrac{b^2}{a^2}}=\dfrac{\sqrt{3}}{2}\).
【C组—拓展题】
1.已知直线\(2x+y-4=0\)经过椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的右焦点\(F_2\),且与椭圆在第一象限的交点为\(A\),与\(y\)轴的交点为\(B\),\(F_1\)是椭圆的左焦点,且\(|AB|=|AF_1 |\),则椭圆的方程为\(\underline{\quad \quad}\).
2.已知椭圆 \(C: \dfrac{x^2}{4}+\dfrac{y^2}{b^2}=1(0<b<2)\),作倾斜角为 \(\dfrac{3 \pi}{4}\)的直线交椭圆\(C\)于\(A\),\(B\)两点,线段\(AB\)的中点为\(M\),\(O\)为坐标原点 \(\overrightarrow{O M}\)与 \(\overrightarrow{M A}\)的夹角为\(θ\),且\(|\tanθ|=3\),则\(b=\)\(\underline{\quad \quad}\).
3.已知点\(A\),\(D\)分别是椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左顶点和上顶点,椭圆的左右焦点分别是\(F_1\)和\(F_2\),点\(P\)是线段\(AD\)上的动点,如果 \(\overrightarrow{P F}_1 \cdot \overrightarrow{P F_2}\)的最大值\(2\),最小值是 \(-\dfrac{2}{3}\),那么椭圆的\(C\)的标准方程是\(\underline{\quad \quad}\).
参考答案
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答案 \(\dfrac{x^2}{5}+y^2=1\)
解析 由题意直线\(2x+y-4=0\)经过椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的右焦点\(F_2\),
令\(y=0\)可得\(x=2\),所以右焦点\(F_2 (2,0)\),即\(c=2\),左焦点\(F_1 (-2,0)\),
由题意令\(x=0\),可得\(y=4\),所以\(B(0,4)\),所以线段\(BF_1\)的中点\(C(-1,2)\),
直线\(BF_1\)的斜率为 \(\dfrac{4-0}{0-(-2)}=2\),
所以线段\(BF_1\)的中垂线方程为 \(y-2=-\dfrac{1}{2}(x+1)\),即\(x+2y-3=0\),
因为\(|AB|=|AF_1 |\),所以线段\(BF_1\)的中垂线过\(A\)点,
所以\(A\)为 \(\left\{\begin{array}{l} x+2 y-3=0 \\ 2 x+y-4=0 \end{array}\right.\)的交点,解得 \(x=\dfrac{5}{3}, y=\dfrac{2}{3}\),即 \(A\left(\dfrac{5}{3}, \dfrac{2}{3}\right)\),
而\(A\)在椭圆上,所以 \(\left\{\begin{array}{l} \dfrac{25}{9 a^2}+\dfrac{4}{9 b^2}=1 \\ c^2=a^2-b^2 \\ c=2 \end{array}\right.\)解得: \(a^2=5,b^2=1\),
所以椭圆的方程为: \(\dfrac{x^2}{5}+y^2=1\),
故答案为: \(\dfrac{x^2}{5}+y^2=1\). -
答案 \(\sqrt{2}\)
解析 根据题意,设\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),\(M(x_0,y_0 )\),
由于\(A\)、\(B\)两点在椭圆上,则 \(\left\{\begin{array}{l} \dfrac{x_1^2}{4}+\dfrac{y_1^2}{b^2}=1 \\ \dfrac{x_2^2}{4}+\dfrac{y_2^2}{b^2}=1 \end{array}\right.\),
两式作差得 \(\dfrac{\left(x_1-x_2\right)\left(x_1+x_2\right)}{4}+\dfrac{\left(y_1-y_2\right)\left(y_1+y_2\right)}{b^2}=0\).
因为直线\(AB\)的倾斜角为 \(\dfrac{3 \pi}{4}\),则有 \(\dfrac{y_1-y_2}{x_1-x_2}=-1\),所以 \(\dfrac{x_0}{4}-\dfrac{y_0}{b^2}=0\),即 \(\dfrac{y_0}{x_0}=\dfrac{b^2}{4}\);
设直线\(OM\)的倾斜角为\(α\),则 \(\theta=\alpha+\dfrac{\pi}{4}\)或 \(\theta=\dfrac{3 \pi}{4}-\alpha\),
\(\tan \theta=\pm \dfrac{\tan \alpha+1}{1-\tan \alpha}\).
又 \(\tan \alpha=\dfrac{y_0}{x_0}=\dfrac{b^2}{4}\),由 \(\dfrac{\left|\dfrac{b^2}{4}+1\right|}{\left|1-\dfrac{b^2}{4}\right|}=3\),解得\(b^2=2\),即 \(b=\sqrt{2}\). -
答案 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\)
解析 \(∵\)画出图形,如图所示;
\(∴\)直线\(AD\)的方程是 \(\dfrac{x}{-a}+\dfrac{y}{b}=1, \quad x \in[-a, 0]\);
\(\therefore{\overrightarrow{P F_1}}{=}(-c-x,-y)\), \({\overrightarrow{P F_2}}=(c-x,-y)\),
\(\overrightarrow{P F}_1 \cdot \overrightarrow{P F_2}=x^2-c^2+y^2=x^2+y^2-c^2\);
设 \(t=x^2+y^2\),则 \(\sqrt{t}\)表示点\(P\)到原点\(O\)的距离,
\(∴\)当\(P\)在\(A\)点时, \(\sqrt{t}\)最大,
此时 \(\overrightarrow{P F}_1 \cdot \overrightarrow{P F_2}=(-a)^2-0^2-c^2=b^2=2\);
当\(P\)在点\(O\)到直线\(AD\)的距离时, \(\sqrt{t}\)最小,
此时 \(\sqrt{t}=\dfrac{1}{\sqrt{\left(\dfrac{1}{-a}\right)^2+\left(\dfrac{1}{b}\right)^2}}\),
\(\therefore t=\dfrac{a^2 b^2}{a^2+b^2}=\dfrac{2 a^2}{a^2+2}\),
\(\therefore \overrightarrow{P F_1} \cdot \overrightarrow{P F_2}=\dfrac{2 a^2}{a^2+2}-\left(a^2-2\right)=-\dfrac{2}{3}\),
整理得 \(3a^4-8a^2-16=0\),解得 \(a^2=4\)或 \(a^2=-\dfrac{4}{3}\)(舍去);
综上, \(a^2=4,b^2=2\),
故椭圆的方程是 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\).
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