河南省2023-2024学年度第二学期八年级第三次学情分析数学.考卷答案

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河南省2023-2024学年度第二学期八年级第三次学情分析数学.考卷答案试卷答案

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分析(1)利用椭圆$C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$过点$(1,\frac{3}{2})$,且离心率为$\frac{1}{2}$,建立方程,求出a,b,即可椭圆C的方程;
(2)动直线l:y=k(x-4)代入椭圆方程,整理可得(3+4k2)x2-32k2x+64k2-12=0,利用韦达定理,结合斜率公式,即可求直线AD与直线AE的斜率之乘积.
(3)运用韦达定理,结合直线的斜率公式和直线恒过定点的求法,化简整理计算即可得到.

解答解:(1)∵椭圆$C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)$过点$(1,\frac{3}{2})$,且离心率为$\frac{1}{2}$,
∴$\frac{1}{{a}^{2}}+\frac{\frac{9}{4}}{{b}^{2}}$=1,$\frac{{a}^{2}-{b}^{2}}{{a}^{2}}$=$\frac{1}{4}$,
∴a=2,b=$\sqrt{3}$,
∴椭圆C的方程$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{3}$=1;
(2)设D(x1,y1),E(x2,y2),动直线l:y=k(x-4)
代入椭圆方程,整理可得(3+4k2)x2-32k2x+64k2-12=0,
∴x1+x2=$\frac{32{k}^{2}}{3+4{k}^{2}}$,x1x2=$\frac{64{k}^{2}-12}{3+4{k}^{2}}$,
∵A(-2,0),
∴直线AD与直线AE的斜率之乘积=$\frac{{y}_{1}}{{x}_{1}+2}$•$\frac{{y}_{2}}{{x}_{2}+2}$=$\frac{{k}^{2}({x}_{1}-4)({x}_{2}-4)}{({x}_{1}+2)({x}_{2}+2)}$=$\frac{9}{4}$;
(3)F(x1,-y1),E(x2,y2),
∴EF的方程为y+y1=$\frac{{y}_{2}+{y}_{1}}{{x}_{2}-{x}_{1}}$(x-x1),即y+k(x1-4)=$\frac{{y}_{2}+{y}_{1}}{{x}_{2}-{x}_{1}}$(x-x1),
令y=0,可得k(x1-4)(x2-x1)=k(x1+x2-8)(x-x1),
∴(x1+x2-8)x=2x1x2-4(x1+x2),
∴x=1,
∴直线EF过定点(1,0).

点评本题考查椭圆方程和性质,主要考查椭圆的离心率公式和方程的运用,联立直线方程,运用韦达定理,同时考查直线恒过定点的求法,属于中档题.