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2026年高考数学北京-18<-->2026年高考数学北京-20
（15分）已知椭圆$E:\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1(a > b > 0)$的一个顶点为$(2,0)$，离心率为$\dfrac{1}{2}$．过点$A(1,1)$且斜率为$k(k\ne \pm 1)$的直线与椭圆$E$交于不同的两点$B$，$C$．点$B$关于直线$y=x$的对称点为$D$，直线$CD$与$y=x$于点$Q$．设△$ABQ$与△$ACQ$的面积分别为$S_{\triangle ABQ}$与$S_{\triangle ACQ}$．
（Ⅰ）求椭圆$E$的方程；
（Ⅱ）若$\vert S_{\triangle ABQ}-S_{\triangle ACQ}\vert =\dfrac{5}{8}$，求$k$的值．
〖答案〗$(I)\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1$；
（Ⅱ）$k=\pm \dfrac{\sqrt{2}}{4}$．
〖分析〗$(I)$利用顶点坐标及离心率计算即可得；
（Ⅱ）设出直线$l_{BC}$，联立曲线方程可得与交点横坐标有关韦达定理，结合题目所给条件计算可得点$D$、点$Q$坐标，再利用点到直线距离公式与两点间距离公式可表示出$S_{\triangle ABQ}$与$S_{\triangle ACQ}$，结合题目所给条件与韦达定理计算即可得解．
〖解答〗解：$(I)$由题意可得$a=2$，则$e=\dfrac{c}{a}=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}=\dfrac{\sqrt{4-{{b}^{2}}}}{2}=\dfrac{1}{2}$，即$b=\sqrt{3}$，故$E$的方程为$\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1$；
（Ⅱ）由题意可得$l_{BC}:y=k(x-1)+1$，设$B(x_{1}$，$y_{1})$、$C(x_{2}$，$y_{2})$，
由$B$关于直线$y=x$对称的点为$D$，则$D(y_{1}$，$x_{1})$，
联立$\left\{\begin{array}{l}\dfrac{{x}^{2}}{4}+\dfrac{{y}^{2}}{3}=1\\ y=k(x-1)+1\end{array}\right.$，消去$y$得：$(4k^{2}+3)x^{2}-8k(k-1)x+4k^{2}-8k-8=0$，
由$\dfrac{{1}^{2}}{4}+\dfrac{{1}^{2}}{3}=\dfrac{7}{12} < 1$，故$A$在椭圆内部，故△$ > 0$恒成立，有${{x}_{1}}+{{x}_{2}}=\dfrac{8k(k-1)}{4{{k}^{2}}+3}$、${{x}_{1}}{{x}_{2}}=\dfrac{4{{k}^{2}}-8k-8}{4{{k}^{2}}+3}$，
则$y_{1}+y_{2}=k(x_{1}-1)+1+k(x_{2}-1)+1=k(x_{1}+x_{2})-2k+2$，
${{y}_{1}}{{y}_{2}}=[k({{x}_{1}}-1)+1]\cdot [k({{x}_{2}}-1)+1]={{k}^{2}}{{x}_{1}}{{x}_{2}}+(k-{{k}^{2}})({{x}_{1}}+{{x}_{2}})+{{k}^{2}}-2k+1$，
${{l}_{CD}}:y=\dfrac{{{y}_{2}}-{{x}_{1}}}{{{x}_{2}}-{{y}_{1}}}(x-{{y}_{1}})+{{x}_{1}}$，联立$\left\{\begin{array}{l}y=\dfrac{{{y}_{2}}-{{x}_{1}}}{{{x}_{2}}-{{y}_{1}}}(x-{{y}_{1}})+{{x}_{1}}\\ y=x\end{array}\right.$，
则$x=\dfrac{{{y}_{2}}-{{x}_{1}}}{{{x}_{2}}-{{y}_{1}}}(x-{{y}_{1}})+{{x}_{1}}$，即$(x_{2}-y_{1})x=(y_{2}-x_{1})x-(y_{2}-x_{1})y_{1}+x_{1}(x_{2}-y_{1})$，
整理得$x=\dfrac{{{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}}}{({{x}_{1}}+{{x}_{2}})-({{y}_{1}}+{{y}_{2}})}$，即$Q(\dfrac{{{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}}}{({{x}_{1}}+{{x}_{2}})-({{y}_{1}}+{{y}_{2}})},\dfrac{{{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}}}{({{x}_{1}}+{{x}_{2}})-({{y}_{1}}+{{y}_{2}})})$，
点$B$到直线$AQ$的距离${{d}_{1}}=\dfrac{\vert {{x}_{1}}-{{y}_{1}}\vert }{\sqrt{2}}$，点$C$到直线$AQ$的距离${{d}_{2}}=\dfrac{\vert {{x}_{2}}-{{y}_{2}}\vert }{\sqrt{2}}$，
又$\vert AQ\vert =\sqrt{2}\vert {{x}_{Q}}-1\vert$，则${{S}_{\triangle ABQ}}=\dfrac{1}{2}\cdot \sqrt{2}\vert {{x}_{Q}}-1\vert \cdot \dfrac{\vert {{x}_{1}}-{{y}_{1}}\vert }{\sqrt{2}}$，${{S}_{\triangle ACQ}}=\dfrac{1}{2}\cdot \sqrt{2}\vert {{x}_{Q}}-1\vert \cdot \dfrac{\vert {{x}_{2}}-{{y}_{2}}\vert }{\sqrt{2}}$，
故$\vert {{S}_{\triangle ABQ}}-{{S}_{\triangle ACQ}}\vert =\dfrac{\sqrt{2}}{2}\vert {{x}_{Q}}-1\vert \cdot \vert \dfrac{\vert {{x}_{1}}-{{y}_{1}}\vert }{\sqrt{2}}-\dfrac{\vert {{x}_{2}}-{{y}_{2}}\vert }{\sqrt{2}}\vert =\dfrac{1}{2}\vert {{x}_{Q}}-1\vert \cdot \vert \vert {{x}_{1}}-{{y}_{1}}\vert -\vert {{x}_{2}}-{{y}_{2}}\vert \vert$
$=\dfrac{1}{2}\vert \dfrac{{{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}}}{({{x}_{1}}+{{x}_{2}})-({{y}_{1}}+{{y}_{2}})}-1\vert \cdot \vert ({{x}_{1}}+{{x}_{2}})-({{y}_{1}}+{{y}_{2}})\vert =\dfrac{1}{2}\vert ({{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}})-({{x}_{1}}+{{x}_{2}})+({{y}_{1}}+{{y}_{2}})\vert$
$=\dfrac{1}{2}\vert ({{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}})-({{x}_{1}}+{{x}_{2}})+({{y}_{1}}+{{y}_{2}})\vert$
$=\dfrac{1}{2}\vert {{x}_{1}}{{x}_{2}}-{{k}^{2}}{{x}_{1}}{{x}_{2}}-(k-{{k}^{2}})({{x}_{1}}+{{x}_{2}})-{{k}^{2}}+2k-1-({{x}_{1}}+{{x}_{2}})+k({{x}_{1}}+{{x}_{2}})-2k+2\vert$
$=\dfrac{1}{2}\vert (1-{{k}^{2}}){{x}_{1}}{{x}_{2}}-(1-{{k}^{2}})({{x}_{1}}+{{x}_{2}})-{{k}^{2}}+1\vert$
$=\dfrac{1}{2}\vert \dfrac{(1-{{k}^{2}})(4{{k}^{2}}-8k-8)}{4{{k}^{2}}+3}-\dfrac{(1-{{k}^{2}})8k(k-1)}{4{{k}^{2}}+3}-{{k}^{2}}+1\vert$
$=\dfrac{1}{2}\vert \dfrac{4{{k}^{2}}-8k-8-4{{k}^{4}}+8{{k}^{3}}+8{{k}^{2}}}{4{{k}^{2}}+3}-\dfrac{8{{k}^{2}}-8k-8{{k}^{4}}+8{{k}^{3}}}{4{{k}^{2}}+3}+\dfrac{-4{{k}^{4}}-3{{k}^{2}}+4{{k}^{2}}+3}{4{{k}^{2}}+3}\vert$
$=\dfrac{1}{2}\vert \dfrac{5{{k}^{2}}-5}{4{{k}^{2}}+3}\vert =\dfrac{5}{8}$，
即有$\vert 5{{k}^{2}}-5\vert =5{{k}^{2}}+\dfrac{15}{4}$，若$k^{2} > 1$，则$5=\dfrac{15}{4}$，无解，不符；
则$k^{2} < 1$，有$5-5{{k}^{2}}=5{{k}^{2}}+\dfrac{15}{4}$，解得$k=\pm \dfrac{\sqrt{2}}{4}$；
故$k=\pm \dfrac{\sqrt{2}}{4}$．
〖点评〗本题主要考查直线和椭圆的位置关系，属于难题．
2026年高考数学北京-18<-->2026年高考数学北京-20
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