3.1.2 椭圆的简单几何性质(2)
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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3颗星!
基础知识
直线与椭圆的位置关系
设直线\(l:Ax+By+C=0\),椭圆\(C:f(x,y)=0\),把两者方程联立得到方程组,消元\(y(\)或\(x)\)得到一个关于\(x(\)或\(y)\)的一元二次方程\(ax^2+bx+c=0(\)或\(ay^2+by+c=0)\).
\(∆>0⇔\)方程有两个不同的实数解,即直线与椭圆有两个交点\(⇔\)相交;
\(∆=0⇔\)方程有两个相同的实数解,即直线与椭圆有一个交点\(⇔\)相切;
\(∆<0⇔\)方程无实数解,即直线与椭圆无交点\(⇔\)相离.
【例】判断直线\(l:y=x+1\)与椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\)的位置关系.
解析 联立方程 \(\left\{\begin{array}{c}
y=x+1 \\
\dfrac{x^2}{4}+\dfrac{y^2}{2}=1
\end{array}\right.\),消\(y\)得\(3x^2+4x-2=0\),
其判别式\(∆=16+24=40>0\),即直线\(l:y=x+1\)与椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\)相离.
直线与椭圆的弦长公式
直线\(y=kx+b\)与椭圆相交于\(A(x_1,y_1 )\),\(B(x_2,y_2)\),则
\(A B=\sqrt{1+k^2} \cdot\left|x_1-x_2\right|=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=\sqrt{1+k^2} \cdot \dfrac{\sqrt{\Delta}}{|a|}\)
或
\(A B=\sqrt{1+\dfrac{1}{k^2}} \cdot\left|y_1-y_2\right|=\sqrt{1+\dfrac{1}{k^2}} \cdot \sqrt{\left(y_1+y_2\right)^2-4 y_1 y_2}=\sqrt{1+\dfrac{1}{k^2}} \cdot \dfrac{\sqrt{\Delta}}{|a|},\),
(\(∆=b^2-4ac\),注意对公式推导的理解,其本质是两点距离公式)
推导
\(A B=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=\sqrt{\left(x_1-x_2\right)^2+\left(k x_1-k x_2\right)^2}\)
\(=\sqrt{\left(x_1-x_2\right)^2+k^2\left(x_1-x_2\right)^2}=\sqrt{1+k^2} \cdot \sqrt{\left(x_1-x_2\right)^2}\)
\(=\sqrt{1+k^2} \cdot\left|x_1-x_2\right|=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}\)
\(=\sqrt{1+k^2} \cdot \sqrt{\dfrac{b^2}{a^2}-\dfrac{4 c}{a}}=\sqrt{1+k^2} \cdot \sqrt{\dfrac{b^2-4 a c}{a^2}}=\sqrt{1+k^2} \cdot \dfrac{\sqrt{\Delta}}{|a|}\).
【例】直线\(l:y=x+1\)与椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\)相交于点\(A\),\(B\),求线段\(AB\)长度.
解析 设\(A(x_1,y_1)\),\(B(x_2,y_2)\),
联立方程 \(\left\{\begin{array}{c}
y=x+1 \\
\dfrac{x^2}{4}+\dfrac{y^2}{2}=1
\end{array}\right.\),消\(y\)得\(3x^2+4x-2=0\),
则 \(x_1+x_2=-\dfrac{4}{3}\), \(x_1 x_2=-\dfrac{2}{3}\),
则 \(A B=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=\sqrt{2} \cdot \sqrt{\dfrac{16}{9}+\dfrac{8}{3}}=\dfrac{4 \sqrt{5}}{3}\).
中点弦
① 涉及到中点弦问题可用点差法求解,在处理双曲线的中点弦问题要注意检验!
② “点差法”的常见题型:求中点弦方程、求(过定点、平行弦)弦中点轨迹、垂直平分线问题.
椭圆常见的结论
(选学内容,证明可见另一专题《圆锥曲线常见二级结论》)
(1) 点\(P\)处的切线\(PT\)平分\(△PF_1 F_2\)在点\(P\)处的外角.(椭圆的光学性质)
(2) 椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的焦半径公式: \(MF_1=a+ex_0\), \(MF_2=a-ex_0\) ,\(M(x_0,y_0)\).
(3) \(AB\)是椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的不平行于对称轴的弦, \(M(x_0,y_0)\)为\(AB\)的中点,则 \(k_{O M} \cdot k_{A B}=-\dfrac{b^2}{a^2}\) , \(k_{A B}=-\dfrac{b^2 x_0}{a^2 y_0}\).
(4) 已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\),\(O\)为坐标原点,\(P\)、\(Q\)为椭圆上两动点,且\(OP⊥OQ\).
(i) \(\dfrac{1}{O P^2}+\dfrac{1}{O Q^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}\) ; (ii) \(O P^2+O Q^2\)的最大值为 \(\dfrac{4 a^2 b^2}{a^2+b^2}\).
(5) 若\(P_0 (x_0,y_0)\)在椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)上,则过\(P_0\)的椭圆的切线方程是 \(\dfrac{x_0 x}{a^2}+\dfrac{y_0 y}{b^2}=1\).
(6) 过椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)上任意一点\(P\)(不是其顶点)作椭圆切线\(PA\),则有 \(k_{O P}=e^2-1\).
(7) 过直线\(mx+ny=1\)且在椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)外一点\(M\)向椭圆引两条切线,切点分别为\(A\),\(B\),则直线\(AB\)必过定点\(N(ma^2,nb^2)\).
基本方法
【题型1】直线与椭圆的位置关系与弦长问题
【典题1】不论\(k\)为何值,直线\(y=kx+1\)与椭圆 \(\dfrac{x^2}{7}+\dfrac{y^2}{m}=1\)有公共点,则实数\(m\)的范围是( )
A.\((0,1)\) \(\qquad \qquad\) B.\([1,+∞)\) \(\qquad \qquad\) C.\([1,7)∪(7,+∞)\) \(\qquad \qquad\) D.\((0,7)\)
解析 方法1 把直线\(y=kx+1\)代入椭圆 \(\dfrac{x^2}{7}+\dfrac{y^2}{m}=1\),
化为\((m+7k^2 ) x^2+14kx+7-7m=0(m≠7,m>0)\).
∵直线\(y=kx+1\)与椭圆 \(\dfrac{x^2}{7}+\dfrac{y^2}{m}=1\)有公共点,
\(∴m+7k^2≠0\),\(△=(14k)^2-4(m+7k^2)(7-7m)≥0\)恒成立.
化为\(1-m≤7k^2\).上式对于任意实数\(k\)都成立,\(∴1-m≤0\),解得\(m≥1\).
\(∴\)实数\(m\)的范围是\([1,7)∪(7,+∞)\).
故选:\(C\).
方法2 从所给含参直线\(y=kx+1\)入手可知直线过定点\((0,1)\),
所以若过定点的直线均与椭圆有公共点,则该点位于椭圆的内部或椭圆上,
所以代入\((0,1)\)后 \(\dfrac{x^2}{7}+\dfrac{y^2}{m} \leq 1\),即 \(\dfrac{1}{m^2}<=1 \Rightarrow m \geq 1\),
因为是椭圆,所以\(m≠7\),故\(m\)的取值范围是\([1,7)∪(7,+∞)\),
答案:\(C\).
点拨 判断直线与椭圆的位置关系可联立方程由判别式确定,也可从几何角度(直线是否过椭圆内一点).
【典题2】已知椭圆\(4x^2+y^2=1\)及直线\(y=x+m\).
(1)当直线和椭圆有公共点时,求实数m的取值范围;
(2)求被椭圆截得的最长弦所在的直线方程.
解析 (1)由 \(\left\{\begin{array}{l}
4 x^2+y^2=1 \\
y=x+m
\end{array}\right.\)得\(5x^2+2mx+m^2-1=0\).
因为直线与椭圆有公共点,
所以\(Δ=4m^2-20(m^2-1)≥0\).解得 \(-\dfrac{\sqrt{5}}{2} \leq m \leq \dfrac{\sqrt{5}}{2}\).
(2)设直线与椭圆交于\(A(x_1,y_1 )\),\(B(x_2,y_2 )\)两点.
由(1)知,\(x_1\),\(x_2\)为方程\(5x^2+2mx+m^2-1=0\)的两根.
由根与系数的关系,得 \(x_1+x_2=-\dfrac{2 m}{5}\), \(x_1 x_2=\dfrac{1}{5}\left(m^2-1\right)\).
所以 \(A B=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=\sqrt{2\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]}\)
\(=\sqrt{2\left[\dfrac{4 m^2}{25}-\dfrac{4}{5}\left(m^2-1\right)\right]}=\dfrac{2}{5} \sqrt{10-8 m^2}\).
所以当\(m=0\)时,\(d\)最大,此时直线方程为\(y=x\).
点拨
1.求直线与椭圆的弦长可使用弦长公式 \(A B=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}\),但要理解其公式的推导.
2.对于韦达定理使用的前提是\(∆≥0\),所以本题第一问中\(m\)的取值范围 \(-\dfrac{\sqrt{5}}{2} \leq m \leq \dfrac{\sqrt{5}}{2}\)对第二问最值很重要.
【典题3】已知椭圆\(C\)的对称中心为坐标原点\(O\),焦点在\(x\)轴上,左、右焦点分别为\(F_1 (-1,0)\)和\(F_2 (1,0)\),点 \(\left(\sqrt{3}, \dfrac{\sqrt{3}}{2}\right)\)在该椭圆上.
(1)求椭圆\(C\)的方程;
(2)过\(F_1\)的直线\(l\)与椭圆\(C\)相交于\(A\),\(B\)两点,若\(△AF_2 B\)的面积为 \(\dfrac{3 \sqrt{5}}{4}\),求以\(F_2\)为圆心且与直线\(l\)相切的圆的方程.
解析 (1)设椭圆的方程: \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\),\(c=1\),即\(a^2-b^2=1\),
将 \(\left(\sqrt{3}, \dfrac{\sqrt{3}}{2}\right)\)代入椭圆方程: \(\dfrac{3}{a^2}+\dfrac{3}{4 b^2}=1\),解得\(a^2=4\),\(b^2=3\),
\(∴\)椭圆\(C\)的方程: \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\);
(2)方法1 当直线\(l\)的斜率不存在时, \(S_{\triangle A F_2 B}=3 \neq \dfrac{3 \sqrt{5}}{4}\),不符合题意,
可设直线方程为:\(y=k(x+1)\),
由 \(\left\{\begin{array}{c}
y=k(x+1) \\
\dfrac{x^2}{4}+\dfrac{y^2}{3}=1
\end{array}\right.\)可得\((4k^2+3) x^2+8k^2 x+4k^2-12=0\),
设\(A(x_1,y_1)\),\(B(x_2,y_2)\),则 \(x_1+x_2=-\dfrac{8 k^2}{4 k^2+3}\), \(x_1 x_2=\dfrac{4 k^2-12}{4 k^2+3}\),
\(\therefore A B=\sqrt{1+k^2} \cdot \dfrac{\sqrt{\Delta}}{|a|}\)\(=\sqrt{1+k^2} \cdot \dfrac{\sqrt{\left(8 k^2\right)^2-4\left(4 k^2+3\right)\left(4 k^2-12\right)}}{4 k^2+3}=12 \cdot \dfrac{1+k^2}{4 k^2+3}\),
又点\(F_2 (1,0)\)到直线\(y=k(x+1)\)的距离为 \(d=\dfrac{|2 k|}{\sqrt{1+k^2}}\),
\(\therefore S_{\triangle A F_2 B}=\dfrac{1}{2} \cdot A B \cdot d=\dfrac{1}{2} \cdot 12 \dfrac{1+k^2}{4 k^2+3} \cdot \dfrac{|2 k|}{\sqrt{1+k^2}}=12 \cdot \dfrac{|k| \sqrt{1+k^2}}{4 k^2+3}=\dfrac{3 \sqrt{5}}{4}\),
解得 \(k=\pm \dfrac{1}{2}\),
则 \(d=\dfrac{|2 k|}{\sqrt{1+k^2}}=\dfrac{2 \sqrt{5}}{5}\),
圆的方程为: \((x-1)^2+y^2=\dfrac{4}{5}\),
以\(F_2\)为圆心且与直线\(l\)相切的圆的方程: \((x-1)^2+y^2=\dfrac{4}{5}\).
方法2 依题意设直线\(l\)方程为:\(x=ty-1\),
联立直线\(l\)与椭圆方程,消去\(x\)可知:\((4+3t^2 ) y^2-6ty-9=0\),
设\(A(x_1,y_1)\),\(B(x_2,y_2)\),则 \(y_1+y_2=\dfrac{6 t}{4+3 t^2}\), \(y_1 y_2=-\dfrac{9}{4+3 t^2}\),
\(\therefore\left|y_1-y_2\right|=\sqrt{\left(y_1+y_2\right)^2-4 y_1 y_2}\)\(=\sqrt{\left(\dfrac{6 t}{4+3 t^2}\right)^2-4 \times\left(-\dfrac{9}{4+3 t^2}\right)}=\dfrac{12 \sqrt{1+t^2}}{4+3 t^2}\),
\(∴△AF_2 B\)的面积 \(S=\dfrac{1}{2}\left|F_1 F_2\right|\left|y_1-y_2\right|=\dfrac{3 \sqrt{5}}{4}\),
即 \(\dfrac{1}{2} \cdot 2 \cdot \dfrac{12 \sqrt{1+t^2}}{4+3 t^2}=\dfrac{3 \sqrt{5}}{4}\),解得:\(t^2=4\),
\(∴t=±2\),则\(F_2\)到直线l的距离 \(d=\dfrac{|-1-1|}{\sqrt{1+t^2}}=\dfrac{2 \sqrt{5}}{5}\),
圆的方程为: \((x-1)^2+y^2=\dfrac{4}{5}\),
以\(F_2\)为圆心且与直线\(l\)相切的圆的方程: \((x-1)^2+y^2=\dfrac{4}{5}\).
点拨
1.本题中设直线方程为\(y=k(x+1)\)要注意斜率是否存在,设直线方程为\(x=ty-1\)避免分类讨论;
2.求三角形面积可以用 \(S_{\triangle A F_2 B}=\dfrac{1}{2} \cdot A B \cdot d\),其中\(AB\)为弦长,\(d\)为\(F_2\)到直线\(AB\)的距离;或用 \(S_{\triangle A F_2 B}=\dfrac{1}{2}\left|F_1 F_2\right|\left|y_1-y_2\right|\).
巩固练习
1.直线\(y=kx+1\)和曲线\(2x^2+3y^2=6\)的位置关系为( )
A.相离 \(\qquad \qquad\) B.相切 \(\qquad \qquad\) C.相交 \(\qquad \qquad\) D.相交或相切
2.已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的离心率为 \(\dfrac{\sqrt{2}}{2}\),且经过点\((\sqrt{2},1)\),直线\(l\)经过\(P(0,1)\),且与椭圆\(C\)相交于\(A\)、\(B\)两点.
(1)求椭圆\(C\)的标准方程;(2)当\(AB=3\),求此时直线\(l\)的方程;
3.椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)过点 \(A\left(1, \dfrac{3}{2}\right)\),离心率为 \(\dfrac{1}{2}\),左、右焦点分别为\(F_1\),\(F_2\),过\(F_1\)的直线交椭圆于\(C\),\(D\)两点.
(1)求椭圆\(C\)的方程;
(2)当\(△F_2 CD\)的面积为 \(\dfrac{12 \sqrt{2}}{7}\)时,求直线的方程.
参考答案
-
答案 \(C\)
解析 曲线\(2x^2+3y^2=6\)为: \(\dfrac{x^2}{3}+\dfrac{y^2}{2}=1\),可得 \(a=\sqrt{3}\), \(b=\sqrt{2}\),
直线\(y=kx+1\)恒过\((0,1)\),定点\((0,1)\)在椭圆内部,
所以直线\(y=kx+1\)与椭圆\(2x^2+3y^2=6\)的位置关系为相交;
故选:\(C\). -
答案 (1) \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\) (2) \(y=\dfrac{\sqrt{2}}{2} x+1\)或 \(y=-\dfrac{\sqrt{2}}{2} x+1\)
解析 (1)椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的离心率为 \(\dfrac{\sqrt{2}}{2}\),
\(\therefore e=\dfrac{c}{a}=\sqrt{1-\dfrac{b^2}{a^2}}=\dfrac{\sqrt{2}}{2}\),\(∴a^2=2b^2\),
\(∵\)椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)经过点 \((\sqrt{2}, 1)\),
\(\therefore \dfrac{2}{a^2}+\dfrac{1}{b^2}=1\),解得\(a^2=4,b^2=2\),
\(∴\)椭圆方程为 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\),
(2)当直线\(l\)的斜率不存在时,即直线\(l\)的方程\(x=0\),
此时 \(A B=2 b=2 \sqrt{2} \neq 3\),
\(∴\)直线\(AB\)的斜率存在,
不妨设直线\(l\)的方程为\(y=kx+1\),设\(A(x_1,y_1)\),\(B(x_2,y_2)\)
联立方程组可得 \(\left\{\begin{array}{l} y=k x+1 \\ x^2+2 y^2=4 \end{array}\right.\),消\(y\)可得\((1+2k^2 ) x^2+4kx-2=0\),
\(\therefore x_1+x_2=-\dfrac{4 k}{1+2 k^2}\), \(x_1 x_2=-\dfrac{2}{1+2 k^2}\),
\(\therefore|A B|=\sqrt{1+k^2} \cdot \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}\)\(=\sqrt{1+k^2} \cdot \sqrt{\left(-\dfrac{4 k}{1+2 k^2}\right)^2+\dfrac{8}{1+2 k^2}}\)
\(=\sqrt{1+k^2} \cdot \sqrt{\dfrac{8+32 k^2}{\left(1+2 k^2\right)^2}}=3\),
整理可得\(4k^4-4k^2+1=0\),解得 \(k^2=\dfrac{1}{2}\),即 \(k=\pm\dfrac{\sqrt{2}}{2}\),
此时直线方程为\(y=\dfrac{\sqrt{2}}{2} x+1\)或 \(y=-\dfrac{\sqrt{2}}{2} x+1\). -
答案 (1) \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\) (2)\(x-y+1=0\)或\(x+y+1=0\)
解析 (1)∵椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)过点 \(A\left(1, \dfrac{3}{2}\right)\),
\(\therefore \dfrac{1}{a^2}+\dfrac{9}{4 b^2}=1\)①,
又\(∵\)离心率为\(\dfrac{1}{2}\), \(\therefore \dfrac{c}{a}=\dfrac{1}{2}\), \(\therefore \dfrac{b^2}{a^2}=\dfrac{3}{4}\) ②,
联立①②得\(a^2=4\),\(b^2=3\).
\(∴\)椭圆的方程为:\(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\);
(2)①当直线的倾斜角为 \(\dfrac{\pi}{2}\)时,取 \(C\left(-1, \dfrac{3}{2}\right)\), \(D\left(-1,-\dfrac{3}{2}\right)\).
\(S_{\triangle A B F_2}=\dfrac{1}{2}|C D| \cdot\left|F_1 F_2\right|=\dfrac{1}{2} \times 3 \times 2 \neq \dfrac{12 \sqrt{2}}{7}\),不适合题意.
②当直线的倾斜角不为 \(\dfrac{\pi}{2}\)时,设直线方程\(l:y=k(x+1)\),
代入 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)得:\((4k^2+3) x^2+8k^2 x+4k^2-12=0\)
设\(C(x_1,y_1)\),\(D(x_2,y_2)\),则\(x_1+x_2=\dfrac{-8 k^2}{4 k^2+3}\) ,\(x_1 x_2=\dfrac{4 k^2-12}{4 k^2+3}\),
\(\therefore|C D|=\sqrt{\left(1+k^2\right)\left[\left(x_1+x_2\right)^2-4 x_1 x_2\right]}\)\(=\sqrt{\left(1+k^2\right)\left[\dfrac{64 k^4}{\left(4 k^2+3\right)^2}-\dfrac{4\left(4 k^2-12\right)}{4 k^2+3}\right.}=\dfrac{12\left(1+k^2\right)}{4 k^2+3}\).
点\(F_2\)到直线\(l\)的距离 \(d=\dfrac{|k+k|}{\sqrt{1+k^2}}\),
\(\therefore S_{\triangle A B F_2}=\dfrac{1}{2}|A B| \cdot d=\dfrac{12|k| \sqrt{1+k^2}}{4 k^2+3}=\dfrac{12 \sqrt{2}}{7}\),
化为\(17k^4+k^2-18=0\),解得\(k^2=1\),\(∴k=±1\),
∴直线方程为:\(x-y+1=0\)或\(x+y+1=0\).
【题型2】直线与椭圆的综合题型
【典题1】动点\(M(x,y)\)与定点\(F(4,0)\)的距离和\(M\)到定直线 \(l: x=\dfrac{25}{4}\)的距离的比是常数 \(\dfrac{4}{5}\),求动点\(M\)的轨迹.
解析 如图,设\(d\)是点\(M\)直线 \(l: x=\dfrac{25}{4}\)的距离,
根据题意,得 \(\dfrac{\sqrt{(x-4)^2+y^2}}{\left|\dfrac{25}{4}-x\right|}=\dfrac{4}{5}\),化简得\(9x^2+25y^2=225\),即 \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\),
所以动点\(M\)的轨迹是长轴、短轴长分别为\(10\),\(6\)的椭圆.
点拨 这是椭圆的第二定义,其中直线 \(l: x=\dfrac{25}{4}\)叫做准线,距离之比\(\dfrac{4}{5}\)为椭圆离心率.
【典题2】椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)上的点到直线\(2 x+\sqrt{3} y-9=0\)的距离的最大值为\(\underline{\quad \quad}\) .
解析 方法1
设与直线 \(2 x+\sqrt{3} y-9=0\)平行的直线\(2x+\sqrt{3} y+m=0\)与椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)相切,
由 \(\left\{\begin{array}{c}
2 x+\sqrt{3} y+m=0 \\
\dfrac{x^2}{4}+\dfrac{y^2}{3}=1
\end{array}\right.\)得\(25x^2+16mx+4m^2-36=0\),
由\(∆=0\)得\(m=±5\),
设直线\(2x+\sqrt{3} y+m=0\)与直线 \(2 x+\sqrt{3} y-9=0\)的距离为\(d\),
当\(m=5\)时, \(d=\dfrac{4 \sqrt{7}}{7}\); 当\(m=-5\)时, \(d=2 \sqrt{7}\).
椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)上的点到直线 \(2 x+\sqrt{3} y-9=0\)的距离的最大值为 \(2 \sqrt{7}\).
方法2
椭圆\(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)的参数方程为 \(\left\{\begin{array}{l}
x=2 \cos \theta \\
y=\sqrt{3} \sin \theta
\end{array}\right.\),\(θ\)为参数,
设椭圆上的动点 \(P(2 \cos \theta, \sqrt{3} \sin \theta)\),则点\(P\)到直线 \(2 x+\sqrt{3} y-9=0\)距离
\(d=\dfrac{|4 \cos \theta+3 \sin \theta-9|}{\sqrt{4+3}}=\dfrac{|5 \sin (\theta+\alpha)-9|}{\sqrt{7}} \leq \dfrac{14}{\sqrt{7}}=2 \sqrt{7}\),其中 \(\tan \alpha=\dfrac{4}{3}\),
\(\therefore d_{\max }=2 \sqrt{7}\).
椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)上的点到直线 \(2 x+\sqrt{3} y-9=0\)的距离的最大值为 \(2 \sqrt{7}\).
【典题3】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\(F_1\)、\(F_2\),离心率为 \(\dfrac{\sqrt{2}}{2}\),过焦点\(F_2\)的直线\(l\)交椭圆\(C\)于\(A\)、\(B\)两点,当直线\(l\)垂直于\(x\)轴时, \(|A B|=\sqrt{2}\).
(1)求椭圆\(C\)的标准方程:
(2)已知椭圆上顶点为\(P\),若直线\(l\)斜率为 \(k=\dfrac{1}{3}\),求证:以\(AB\)为直径的圆过点\(P\).
解析 (1)解:设椭圆的焦距为\(2c\),当直线\(l\)垂直于\(x\)轴时,\(l\)的方程为\(x=c\),
则\(A\),\(B\)的坐标为 \(\left(c, \dfrac{b^2}{a}\right)\), \(\left(c,-\dfrac{b^2}{a}\right)\),则 \(|A B|=\dfrac{2 b^2}{a}=\sqrt{2}\),
又 \(\dfrac{c}{a}=\dfrac{\sqrt{2}}{2}\),\(a^2=b^2+c^2\),解得 \(a=\sqrt{2}\),\(b=1\).
\(∴\)椭圆方程为 \(\dfrac{x^2}{2}+y^2=1\);
(2)证明:由(1)可知,\(P(0,1)\),直线\(l\)的方程为 \(y=\dfrac{1}{3} x-\dfrac{1}{3}\).
设 \(A\left(x_1, y_1\right)\), \(B\left(x_2, y_2\right)\),
联立 \(\left\{\begin{array}{l}
y=\dfrac{1}{3} x-\dfrac{1}{3} \\
\dfrac{x^2}{2}+y^2=1
\end{array}\right.\),消去\(y\)可得 \(\dfrac{2}{18} x^2-\dfrac{2}{9} x-\dfrac{8}{9}=0\).
则有 \(x_1+x_2=\dfrac{4}{2}\), \(x_1 x_2=-\dfrac{16}{2}\).
\(\therefore \overrightarrow{P A} \cdot \overrightarrow{P B}=x_1 \cdot x_2+\left(y_1-1\right)\left(y_2-1\right)=x_1 \cdot x_2+\left(\dfrac{1}{3} x_1-\dfrac{1}{3}-1\right)\left(\dfrac{1}{3} x_2-\dfrac{1}{3}-1\right)\)
\(=\dfrac{10}{9} x_1 x_2-\dfrac{4}{9}\left(x_1+x_2\right)+\dfrac{16}{9}=\dfrac{10}{9} \times\left(-\dfrac{16}{2}\right)-\dfrac{4}{9} \times \dfrac{4}{2}+\dfrac{16}{9}=0\).
\(∴\)点\(P\)在以\(AB\)为直径的圆上.
巩固练习
1.过椭圆 \(\dfrac{x^2}{16}+\dfrac{y^2}{4}=1\)内一点\(M(2,1)\)引一条弦,使弦被点\(M\)平分,求这条弦所在的直线方程.
2.已知椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\)的左焦点为\(F\),左、右顶点分别为\(A\)、\(B\).椭圆上任一点到直线\(l:x=m\)的距离与点\(F\)的距离之比为\(2\).
(1)求\(m\)
(2)若斜率不为\(0\)且过\(F\)的直线与椭圆交于\(C\),\(D\)两点,过\(B\),\(C\)的直线与\(l\)交于点\(M\),证明:\(M\),\(A\),\(D\)三点共线.
3.椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的离心率为 \(\dfrac{3}{5}\),\(P(m,0)\)为\(C\)的长轴上的一个动点,过\(P\)点斜率为 \(\dfrac{4}{5}\)的直线\(l\)交\(C\)于\(A\)、\(B\)两点.当\(m=0\)时, \(\overrightarrow{P A} \cdot \overrightarrow{P B}=-\dfrac{41}{2}\).
(1)求\(C\)的方程;(2)求证:\(|PA|^2+|PB|^2\)为定值.
参考答案
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答案 \(x+2y-4=0\)
解析 设所求直线方程为\(y-1=k(x-2)\),代入椭圆方程并整理得:
\((4k^2+1) x^2-8(2k^2-k)x+4(2k-1)^2-16=0\)
又设直线与椭圆的交点为 \(A\left(x_1, y_1\right)\), \(B\left(x_2, y_2\right)\),
则\(x_1\),\(x_2\)是方程的两个根,于是 \(x_1+x_2=\dfrac{8\left(2 k^2-k\right)}{4 k^2+1}\),
又\(M\)为\(AB\)的中点,所以 \(\dfrac{x_1+x_2}{2}=\dfrac{4\left(2 k^2-k\right)}{4 k^2+1}=2\),解得 \(k=-\dfrac{1}{2}\),
故所求直线方程为\(x+2y-4=0\). -
答案 (1)\(m=-4\) (2)略
解析 (1)椭圆左焦点\(F\)的坐标为\((-1,0)\),\(A\),\(B\)坐标分别为\((-2,0)\),\((2,0)\),
设椭圆上任意一点为\(P(x_0,y_0)\),则 \(\dfrac{\left|x_0-m\right|}{\sqrt{\left(x_0+1\right)^2+y_0^2}}=2\),而 \(\dfrac{x_0{ }^2}{4}+\dfrac{y_0^2}{3}=1 \text {, }\),
整理得\(|x_0-m|=|x_0+4|\)对任意\(-2≤x_0≤2\)恒成立,
\(∴m=-4\).
(2)设\(C\),\(D\)两点的坐标分别为\((x_1,y_1 )\),\((x_2,y_2 )\),过\(F\)的直线为\(x=ny-1\),
由 \(\left\{\begin{array}{l} \dfrac{x^2}{4}+\dfrac{y^2}{3}=1 \\ x=n y-1 \end{array}\right.\)消去\(x\)得:\((3n^2+4) y^2-6ny-9=0\),
则 \(y_1+y_2=\dfrac{6 n}{3 n^2+4}\), \(y_1 y_2=-\dfrac{9}{3 n^2+4}\),
直线\(BC\)的方程为 \(y=\dfrac{y_1}{x_1-2}(x-2)\),其与\(l\)的交点\(M\)的坐标为 \(\left(-4,-\dfrac{6 y_1}{x_1-2}\right)\),
故 \(\overrightarrow{A M}=\left(-2,-\dfrac{6 y_1}{x_1-2}\right)\), \(\overrightarrow{A D}=\left(x_2+2, y_2\right)\),
由于\(x_1=ny_1-1\),\(x_2=ny_2-1\),
则 \(\overrightarrow{A M}=\left(-2,-\dfrac{6 y_1}{n y_1-3}\right)\), \(\overrightarrow{A D}=\left(n y_2+1, y_2\right)\),
由于 \(-2 \times y_2+\dfrac{6 y_1}{n y_1-3} \times\left(n y_2+1\right)=\dfrac{4 n y_1 y_2+6\left(y_1+y_2\right)}{n y_1-3}=0\),
故 \(\overrightarrow{A M} / / \overrightarrow{A D}\),即得\(M\),\(A\),\(D\)三点共线. -
答案 (1) \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\) (2)略
解析 (1)因为离心率为 \(\dfrac{3}{5}\),所以 \(\dfrac{b}{a}=\dfrac{4}{5}\).
当\(m=0\)时,\(l\)的方程为 \(y=\dfrac{4}{5} x\),
代入\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\),并整理得 \(x^2=\dfrac{a^2}{2}\).
设\(A(x_0,y_0)\),则\(B(-x_0,-y_0)\),\(P(m,0)\),
\(\overrightarrow{P A} \cdot \overrightarrow{P B}=-x_0^2-y_0^2=-\dfrac{41}{25} x_0^2=-\dfrac{41}{25} \cdot \dfrac{a^2}{2}\).
又因为 \(\overrightarrow{P A} \cdot \overrightarrow{P B}=-\dfrac{41}{2}\),所以\(a^2=25\),\(b^2=16\),
椭圆\(C\)的方程为 \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\).
(2)\(l\)的方程为 \(x=\dfrac{5}{4} y+m\),代入 \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\),
并整理得\(25y^2+20my+8(m^2-25)=0\).
设\(A(x_1,y_1)\),\(B(x_2,y_2)\),
则 \(|P A|^2=\left(x_1-m\right)^2+y_1^2=\dfrac{41}{16} y_1^2\),同理 \(|P B|^2=\dfrac{41}{16} y_2^2\).
则 \(|P A|^2+|P B|^2=\dfrac{41}{16}\left(y_1^2+y_2^2\right)=\dfrac{41}{16}\left[\left(y_1+y_2\right)^2-2 y_1 y_2\right]\)
\(=\dfrac{41}{16}\left[\left(-\dfrac{4 m}{5}\right)^2-\dfrac{16\left(m^2-25\right)}{25}\right]=41\).
所以\(|PA|^2+|PB|^2\)是定值.
分层练习
【A组---基础题】
1.若过椭圆 \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=1\)内一点\(P(2,1)\)的弦被该点平分,则该弦所在的直线方程为( )
A.\(8x+9y-25=0\) \(\qquad \qquad\) B.\(3x-4y-5=0\) \(\qquad \qquad\) C.\(4x+3y-15=0\) \(\qquad \qquad\) D.\(4x-3y-9=0\)
2.椭圆 \(C: \dfrac{x^2}{9}+\dfrac{y^2}{4}=1\)上的点\(P\)到直线\(l:4x+3y+18=0\)的距离的最小值为( )
A. \(\dfrac{18-6 \sqrt{5}}{5}\) \(\qquad \qquad\) B. \(\dfrac{18-\sqrt{5}}{5}\) \(\qquad \qquad\) C. \(\dfrac{18+\sqrt{5}}{5}\) \(\qquad \qquad\) D. \(\dfrac{18+6 \sqrt{5}}{5}\)
3.直线\(y=x+1\)被椭圆 \(\dfrac{x^2}{4}+\dfrac{y^2}{2}=1\)所截得线段的中点的坐标是\(\underline{\quad \quad}\) .
4.椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的左顶点到右焦点的距离为 \(\sqrt{3}+\sqrt{2}\),椭圆上的点到右焦点的距离的最小值为 \(\sqrt{3}-\sqrt{2}\).
(1)求椭圆\(C\)的方程;
(2)设斜率为\(1\)的直线\(l\)经过椭圆上顶点,并与椭圆交于\(A\),\(B\)两点,求\(|AB|\).
5.已知椭圆 \(\dfrac{x^2}{36}+\dfrac{y^2}{9}=1\)和点\(P(4,2)\),直线\(l\)经过点\(P\)且与椭圆交于\(A\),\(B\)两点.
(1)当直线\(l\)的斜率为 \(\dfrac{1}{2}\)时,求线段\(AB\)的长度;
(2)当P点恰好为线段AB的中点时,求\(l\)的方程.
6.已知椭圆 \(C: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的短轴长为 \(4 \sqrt{2}\),离心率为 \(\dfrac{1}{3}\).
(1)求椭圆\(C\)的标准方程;
(2)设椭圆\(C\)的左,右焦点分别为\(F_1\),\(F_2\),左,右顶点分别为\(A\),\(B\),点\(M\),\(N\)为椭圆\(C\)上位于\(x\)轴上方的两点,且\(F_1 M∥F_2 N\),直线\(F_1 M\)的斜率为 \(2 \sqrt{6}\),记直线\(AM\),\(BN\)的斜率分别为\(k_1\),\(k_2\),求\(3k_1+2k_2\)的值.
参考答案
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答案 \(A\)
解析 设弦的两端点为\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),\(P\)为\(AB\)中点,
\(A\),\(B\)在椭圆上, \(\dfrac{x_1{ }^2}{9}+\dfrac{y_1{ }^2}{4}=1\), \(\dfrac{x_2{ }^2}{9}+\dfrac{y_2{ }^2}{4}=1\),
两式相减得: \(\dfrac{x_1^2-x_2^2}{9}+\dfrac{y_1^2-y_2^2}{4}=0\),
由\(x_1+x_2=4\),\(y_1+y_2=2\),可得: \(\dfrac{y_1-y_2}{x_1-x_2}=-\dfrac{8}{9}\),
则 \(k=-\dfrac{8}{9}\),且过点\(P(2,1)\),有 \(y-1=-\dfrac{8}{9}(x-2)\),
整理得\(8x+9y-25=0\).
故选:\(A\). -
答案 \(A\)
解析 设点\(P\)的坐标为 \((3 \cos \theta, 2 \sin \theta)\),其中\(θ∈[0,2π)\),
则点\(P\)到直线\(l\)的距离 \(d=\dfrac{|12 \cos \theta+6 \sin \theta+18|}{\sqrt{4^2+3^2}}=\dfrac{6|2 \cos \theta+\sin \theta+3|}{5}\)
\(=\dfrac{6}{5}|\sqrt{5} \sin (\theta+\beta)+3| \geq \dfrac{6|3-\sqrt{5}|}{5}\),其中\(\tanβ=2\),
当\(\sin(θ+β)=-1\)时,等号成立.
\(d\)取得最小值 \(\dfrac{18-6 \sqrt{5}}{5}\).
故选:\(A\). -
答案 \(\left(-\dfrac{2}{3}, \dfrac{1}{3}\right)\)
解析 联立方程 \(\left\{\begin{array}{l} y=x+1 \\ \dfrac{x^2}{4}+\dfrac{y^2}{2}=1 \end{array}\right.\),消去\(y\)得\(3x^2+4x-2=0\).
设交点为\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),中点\(M(x_0,y_0)\),
\(\therefore x_1+x_2=-\dfrac{4}{3}\), \(x_0=\dfrac{x_1+x_2}{2}=-\dfrac{2}{3}\), \(y_0=x_0+1=\dfrac{1}{3}\).
\(∴\)所求中点的坐标为 \(\left(-\dfrac{2}{3}, \dfrac{1}{3}\right)\). -
答案 (1) \(\dfrac{x^2}{3}+y^2=1\) (2) \(\dfrac{3 \sqrt{2}}{2}\)
解析 (1)设椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的半焦距为\(c\),
由题意可得 \(\left\{\begin{array}{l} a+c=\sqrt{3}+\sqrt{2} \\ a-c=\sqrt{3}-\sqrt{2} \end{array}\right.\),解得 \(a=\sqrt{3}\), \(c=\sqrt{2}\).
\(∴b^2=a^2-c^2=1\).
则椭圆\(C\)的方程为 \(\dfrac{x^2}{3}+y^2=1\);
(2)如图,椭圆\(C\)的上顶点\(A(0,1)\),
则直线\(l\)的方程\(y=x+1\).
联立 \(\left\{\begin{array}{l} y=x+1 \\ \dfrac{x^2}{3}+y^2=1 \end{array}\right.\),得\(2x^2+3x=0\).解得:\(x_A=0\), \(x_B=-\dfrac{3}{2}\).
\(\therefore|A B|=\sqrt{2}\left|x_A-x_B\right|=\dfrac{3 \sqrt{2}}{2}\).
-
答案 (1) \(3 \sqrt{10}\) (2) \(y=-\dfrac{1}{2} x+4\)
解析 (1)由已知可得直线\(l\)的方程为 \(y-2=\dfrac{1}{2}(x-4)\),
即 \(y=\dfrac{1}{2} x\).由 \(\left\{\begin{array}{l} y=\dfrac{1}{2} x \\ \dfrac{x^2}{36}+\dfrac{y^2}{9}=1 \end{array}\right.\),可得\(x^2-18=0\),
若设\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),
则\(x_1+x_2=0\),\(x_1 x_2=-18\).
于是 \(|A B|=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=\sqrt{\left(x_1-x_2\right)^2+\dfrac{1}{4}\left(x_1-x_2\right)^2}\)
\(=\dfrac{\sqrt{5}}{2} \sqrt{\left(x_1+x_2\right)^2-4 x_1 x_2}=\dfrac{\sqrt{5}}{2} \times 6 \sqrt{2}=3 \sqrt{10}\).
所以线段\(AB\)的长度为 \(3 \sqrt{10}\).
(2)设\(A(x_1,y_1 )\),\(B(x_2,y_2 )\),则有 \(\left\{\begin{array}{l} \dfrac{x_1^2}{36}+\dfrac{y_1^2}{9}=1 \\ \dfrac{x_2^2}{36}+\dfrac{y_2^2}{9}=1 \end{array}\right.\),
两式相减得 \(\dfrac{x_2^3-x_1^2}{36}+\dfrac{y_2^2-y_1^2}{9}=0\).
由于\(P(4,2)\)是\(AB\)的中点,
\(∴x_1+x_2=8\),\(y_1+y_2=4\),
从而\((x_2-x_1 )+2(y_2-y_1 )=0\), \(k_{A B}=\dfrac{y_2-y_1}{x_2-x_1}=-\dfrac{1}{2}\),
于是直线\(AB\)的方程为 \(y-2=-\dfrac{1}{2}(x-4)\),即 \(y=-\dfrac{1}{2} x+4\). -
答案 (1) \(\dfrac{x^2}{9}+\dfrac{y^2}{8}=1\) (2) \(0\)
解析 (1)由题意,得 \(2 b=4 \sqrt{2}\), \(\dfrac{c}{a}=\dfrac{1}{3}\),
又\(a^2-c^2=b^2\),\(∴a=3\), \(b=2 \sqrt{2}\),\(c=1\).
\(∴\)椭圆方程为 \(\dfrac{x^2}{9}+\dfrac{y^2}{8}=1\);
(2)由(1),可知\(A(-3,0)\),\(B(3,0)\),\(F_1 (-1,0)\),
据题意,\(F_1 M\)的方程为 \(y=2 \sqrt{6}(x+1)\).
记直线\(F_1 M\)与椭圆的另一交点为\(M'\),设\(M(x_1,y_1 )(y_1>0)\),\(M'(x_2,y_2)\),
\(∵F_1 M∥F_2 N\),根据对称性,得\(N(-x_2,-y_2 )\),
联立 \(\left\{\begin{array}{l} 8 x^2+9 y^2=72 \\ y=2 \sqrt{6}(x+1) \end{array}\right.\),消去\(y\),得\(14x^2+27x+9=0\).
\(∵x_1>x_2\), \(\therefore x_1=-\dfrac{3}{7}\), \(x_2=-\dfrac{3}{2}\),
\(\because k_1=\dfrac{y_1}{x_1+3}=\dfrac{2 \sqrt{6}\left(x_1+1\right)}{x_1+3}=\dfrac{4 \sqrt{6}}{9}\), \(k_2=\dfrac{-y_2}{-x_2-3}=\dfrac{2 \sqrt{6}\left(x_2+1\right)}{x_2+3}=\dfrac{-2 \sqrt{6}}{3}\).
\(\therefore 3 k_1+2 k_2=3 \times \dfrac{4 \sqrt{6}}{9}+2 \times\left(-\dfrac{2 \sqrt{6}}{3}\right)=0\),
即\(3k_1+2k_2\)的值为\(0\).
【B组---提高题】
1.已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>b>0)\)的离心率为 \(\dfrac{\sqrt{3}}{3}\),过椭圆的右焦点且斜率为 \(\dfrac{1}{2}\)的直线与椭圆交于\(A\),\(B\)两点,则\(△AOB\)(其中\(O\)为原点)的形状为( )
A.锐角三角形\(\qquad \qquad\) B.钝角三角形\(\qquad \qquad\) C.直角三角形 \(\qquad \qquad\) D.锐角或直角三角形
2.如图,在平面直角坐标系\(xoy\)中,已知椭圆 \(C_1: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\)和椭圆 \(C_2: \dfrac{x^2}{c^2}+\dfrac{y^2}{b^2}=1\),其中\(a>c>b>0\),\(a^2=b^2+c^2\),\(C_1\),\(C_2\)的离心率分别为\(e_1\),\(e_2\),且满足 \(e_1: e_2=2: \sqrt{3}\),\(A\),\(B\)分别是椭圆\(C_2\)的右、下顶点,直线\(AB\)与椭圆\(C_1\)的另一个交点为\(P\),且 \(P B=\dfrac{18}{5}\).
(1)求椭圆\(C_1\)的方程;
(2)与椭圆\(C_2\)相切的直线\(MN\)交椭圆\(C_1\)与点\(M\),\(N\),求\(MN\)的最大值.
参考答案
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答案 \(B\)
解析 由椭圆的离心率可得 \(\dfrac{\sqrt{a^2-b^2}}{a}=\dfrac{\sqrt{3}}{3}\)可得 \(\sqrt{2} a=\sqrt{3} b\),则椭圆的方程为 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{\dfrac{2}{3} a^2}=1\),
椭圆的右焦点为 \(F\left(\dfrac{\sqrt{3}}{3} a, 0\right)\),由直线\(l\)的方程为 \(y=\dfrac{1}{2}\left(x-\dfrac{\sqrt{3}}{3} a\right)\),
由 \(\left\{\begin{array}{l} \dfrac{x^2}{a^2}+\dfrac{y^2}{\dfrac{2}{3} a^2}=1 \\ y=\dfrac{1}{2}\left(x-\dfrac{\sqrt{3}}{3} a\right) \end{array}\right.\)可得 \(2 x^2-2 \sqrt{3} a x-7 a^2=0\),
设\(A(x_1,y_1)\),\(B(x_2,y_2)\),则 \(\left\{\begin{array}{l} x_1+x_2=\dfrac{2 \sqrt{3}}{2} a \\ x_1 x_2=\dfrac{-7 a^2}{2} \end{array}\right.\),
则 \(y_1 y_2=\dfrac{1}{4}\left(x_1-\dfrac{\sqrt{3}}{3} a\right)\left(x_2-\dfrac{\sqrt{3}}{3} a\right)=\dfrac{1}{4} a\left(x_1+x_2\right)+\dfrac{1}{12} a^2=-\dfrac{8}{33} a^2\),
则 \(\overrightarrow{O A} \cdot \overrightarrow{O B}=x_1 x_2+y_1 y_2=-\dfrac{29}{33} a^2<0\),
\(∴∠AOB\)一定为钝角,
故选:\(B\). -
答案 (1) \(\dfrac{x^2}{9}+\dfrac{y^2}{3}=1\) (2) \(\dfrac{3 \sqrt{2}}{2}\)
解析 (1)由题意知 \(e_1=\dfrac{c}{a}\), \(e_2=\dfrac{\sqrt{c^2-b^2}}{c}=\dfrac{\sqrt{2 c^2-a^2}}{c}\),
因为 \(e_1: e_2=2: \sqrt{3}\),所以 \(\sqrt{3} \cdot \dfrac{c}{a}=2 \cdot \dfrac{\sqrt{2 c^2-a^2}}{c}\),化简得 \(a^2=\dfrac{3}{2} c^2\),
又\(a^2=b^2+c^2\),所以 \(a=\sqrt{3} b\), \(c=\sqrt{2} b\),所以 \(A(\sqrt{2} b, 0)\),\(B(0,-b)\),
所以直线\(AB\)的方程为 \(y=\dfrac{\sqrt{2}}{2} x-b\),
与椭圆 \(C_1: \dfrac{x^2}{3 b^2}+\dfrac{y^2}{b^2}=1\)联立并消去\(y\),得 \(x^2+3\left(\dfrac{\sqrt{2}}{2} x-b\right)^2=3 b^2\),
整理得\(x_1=0\), \(x_2=\dfrac{6 \sqrt{2}}{5} b\),所以 \(P\left(\dfrac{6 \sqrt{2}}{5} b, \dfrac{b}{5}\right)\),
因为 \(P B=\dfrac{18}{5}\),所以 \(\sqrt{\left(\dfrac{6 \sqrt{2}}{5} b-0\right)^2+\left(\dfrac{b}{5}+b\right)^2}=\dfrac{18}{5}\),
得 \(b=\sqrt{3}\),所以\(a=3\),
椭圆\(C_1\)的方程为 \(\dfrac{x^2}{9}+\dfrac{y^2}{3}=1\).
(2)当直线\(MN\)的斜率不存在时,易得\(MN=2\).
当直线\(MN\)的斜率存在时,设直线\(MN:y=kx+m(k≠0)\),
与椭圆 \(C_2: \dfrac{x^2}{6}+\dfrac{y^2}{3}=1\)联立并消去\(y\),
得\((1+2k^2 ) x^2+4knx+2m^2-6=0\),
因为直线\(MN\)与椭圆\(C_2\)相切,所以\(∆=16k^2 m^2-4(1+2k^2 )(2m^2-6)=0\),
整理得\(6k^2+3-m^2=0 (*)\),
将直线\(MN\)与椭圆\(C_1\)方程联立并消去y,得\((1+3k^2 ) x^2+6kmx+3m^2-9=0\),
由\((*)\)式可得\(∆=36k^2 m^2-4(1+3k^2 )(3m^2-9)=12(9k^2+3-m^2 )=36k^2\).
设\(M(x_M,y_M )\),\(N(x_N,y_N)\),则 \(x_M+x_N=-\dfrac{6 k m}{1+3 k^2}\), \(x_M x_N=\dfrac{3 m^2-9}{1+3 k^2}\),
所以 \(M N=\sqrt{1+k^2}\left|x_M-x_N\right|=\sqrt{1+k^2} \cdot \dfrac{\sqrt{36 k^2}}{1+3 k^2}=6 \sqrt{\dfrac{k^4+k^2}{\left(1+3 k^2\right)^2}}\),
设 \(t=1+3 k^2\),则\(t>1\), \(M N=6 \sqrt{\dfrac{t^2+t-2}{9 t^2}}=2 \sqrt{-2\left(\dfrac{1}{t}-\dfrac{1}{4}\right)^2+\dfrac{9}{8}} \leq \dfrac{3 \sqrt{2}}{2}\),
综上所述,所以当\(t=4\),即\(k=±1\)时,\(MN\)最大,且最大值为\(\dfrac{3 \sqrt{2}}{2}\).
【C组---拓展题】
1.直线\(3x+4y=12\)与椭圆 \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\)相交于\(A\),\(B\)两点,该椭圆上点\(P\)使得\(△PAB\)的面积等于\(4\),这样的点\(P\)共有( )
A.\(1\)个 \(\qquad \qquad\) B.\(2\)个 \(\qquad \qquad\) C.\(3\)个 \(\qquad \qquad\) D.\(4\)个
2.如图,在平面直角坐标系\(xOy\)中,已知椭圆 \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1(a>0, b>0)\)的离心率为 \(\dfrac{1}{2}\),且过点 \(\left(1, \dfrac{3}{2}\right)\).\(F\)为椭圆的右焦点,\(A\),\(B\)为椭圆上关于原点对称的两点,连接\(AF\),\(BF\)分别交椭圆于\(C\),\(D\)两点.
(1)求椭圆的标准方程;
(2)若\(AF=FC\),求\(\dfrac{B F}{F D}\)的值;
(3)设直线\(AB\),\(CD\)的斜率分别为\(k_1\),\(k_2\),是否存在实数\(m\),使得\(k_2=mk_1\),若存在,求出\(m\)的值;若不存在,请说明理由.
参考答案
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答案 \(B\)
解析 联立直线直线\(3x+4y=12\)与椭圆 \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\),得 \(\left\{\begin{array}{l} x=4 \\ y=0 \end{array}\right.\)或 \(\left\{\begin{array}{l} x=0 \\ y=3 \end{array}\right.\),
\(∴\)直线与椭圆的交点为\(A(4,0)\)和\(B(0,3)\),得 \(|A B|=\sqrt{4^2+3^2}=5\),
设点\(P\)到\(AB\)的距离为\(d\),则 \(S_{\triangle P A B}=\dfrac{1}{2} \times|A B| \times d=4\),
即 \(\dfrac{1}{2} \times 5 \times d=4\),解之得 \(d=\dfrac{8}{5}\);
再设平行于直线\(3x+4y=12\)与椭圆相切的直线为\(3x+4y+m=0\),
与椭圆 \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\)联立消去x可得:\(32y^2+8my+m^2-144=0\),
\(Δ=64m^2-4×32×(m^2-144)=0\).
可得 \(m=\pm 12 \sqrt{2}\),
由此可得两条平行于直线\(3x+4y=12\)的切线分别为:
\(l_1:3x+4y+12\sqrt{2}=0\)和\(l_2:3x+4y-12\sqrt{2}=0\),
\(∵l_1\)与直线\(3x+4y=12\)的距离 \(d_1=\dfrac{|12-12 \sqrt{2}|}{5}=\dfrac{12}{5}(\sqrt{2}-1)<\dfrac{8}{5}\),
\(l_2\)与直线\(3x+4y=12\)的距离 \(d_2=\dfrac{|12+12 \sqrt{2}|}{5}=\dfrac{12}{5}(\sqrt{2}+1)>\dfrac{8}{5}\),
\(∴l_1\)与\(l_2\)中,\(l_1\)与椭圆相交,有两个交点,而\(l_2\)与椭圆相离,没有交点.
因此有两个\(P\)点使\(△PAB\)的面积等于\(4\),
故选:\(B\). -
答案 (1) \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\) (2) \(\dfrac{7}{3}\) (3) \(m=\dfrac{5}{3}\)
解析 (1)由题意知 \(e=\dfrac{c}{a}=\dfrac{1}{2}\),则\(a=2c\),\(b^2=a^2-c^2=3c^2\),
将 \(\left(1, \dfrac{3}{2}\right)\)代入椭圆方程: \(\dfrac{1}{4 c^2}+\dfrac{9}{3 c^2 \times 4}=1\),解得:\(c=1\),
则\(a=2\), \(b=\sqrt{3}\),
\(∴\)椭圆方程为: \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\);
(2)若\(AF=FC\),由椭圆对称性,知 \(A\left(1, \dfrac{3}{2}\right)\),则 \(B\left(-1,-\dfrac{3}{2}\right)\),\(F(1,0)\)
此时直线\(BF\)方程为\(3x-4y-3=0\),
由 \(\left\{\begin{array}{l} 3 x-4 y-3=0 \\ \dfrac{x^2}{4}+\dfrac{y^2}{3}=1 \end{array}\right.\),整理得\(7x^2-6x-13=0\),解得 \(x=\dfrac{13}{7}\)(\(x=-1\)舍去),
故 \(\dfrac{B F}{F D}=\dfrac{1-(-1)}{\dfrac{13}{7}-1}=\dfrac{7}{3}\).
(3)设\(A(x_0,y_0 )\),则\(B(-x_0,-y_0 )\),
直线\(AF\)的方程为 \(y=\dfrac{y_0}{x_0-1}(x-1)\),代入椭圆方程 \(\dfrac{x^2}{4}+\dfrac{y^2}{3}=1\),
得\((15-6x_0 ) x^2-8y_0^2 x-15x_0^2+24x_0=0\),
即\((15-6x_0 ) x^2-(24-6x_0^2 )x-15x_0^2+24x_0=0\),
因为\(x=x_0\)是该方程的一个解,所以\(C\)点的横坐标 \(x_C=\dfrac{8-5 x_0}{5-2 x_0}\),
又\(C(x_C,y_C )\)在直线 \(y=\dfrac{y_0}{x_0-1}(x-1)\)上,所以 \(y_C=\dfrac{y_0}{x_0-1}\left(x_C-1\right)=\dfrac{-3 y_0}{5-2 x_0}\),
同理,\(D\)点坐标为 \(\left(\dfrac{8+5 x_0}{5+2 x_0}, \dfrac{3 y_0}{5+2 x_0}\right)\) ,
所以 \(k_2=\dfrac{\dfrac{3 y_0}{5+2 x_0}-\dfrac{-3 y_0}{5-2 x_0}}{\dfrac{8+5 x_0}{5+2 x_0}-\dfrac{8-5 x_0}{5-2 x_0}}=\dfrac{5 y_0}{3 x_0}=\dfrac{5}{3} k_1\),
即存在 \(m=\dfrac{5}{3}\),使得\(k_2=\dfrac{5}{3} k_1\).