Question

请教高中数学问题，求高手解答，要有详细步骤哦~

Answer 1

由题意得到e=c/a=1/2, b=2根号3.则有c^2/a^2=1/4, 即有1-b^2/a^2=1/4
12/a^2=3/4, a^2=16
故椭圆方程是x^2/16+y^2/12=1
2.设A(x1,y1),B(x2,y2),直线AB方程是y=1/2x+t代入到椭圆中有:
x^2+tx+t^2-12=0

由判别式=t^2-4(t^2-12)>0,得到-4<t<4
x1+x2=-t,x1x2=t^2-12
x=2代入椭圆中有y=土3,即有PQ=6

故有S(APBQ)=1/2PQ*|x2-x1|=1/2*6*根号[t^2-4(t^2-12)]=3根号(48-3t^2)
故当t=0时,S有最大值是12根号3.
3.P坐标是(2,3),则有k1=K(PA)=(Y1-3)/(x1-2),k2=K(PB)=(y2-3)/(x2-2)
k1+k2=(y1-3)/(x1-2)+(y2-3)/(x2-2)=(1/2x1+t-3)/(x1-2)+(1/2x2+t-3)/(x2-2)
=[(1/2x1+t-3)*(x2-2)+(1/2x2+t-3)(x1-2)]/(x1-2)(x2-2)
=[1/2x1x2-x1+(t-3)x2-2(t-3)+1/2x1x2-x2+(t-3)x1-2(t-3)]/(x1x2-2(x1+x2)+4)
=[x1x2+(t-4)(x1+x2)-4(t-3)]/(t^2-12-2(-t)+4)
=(t^2-12+(t-4)*(-t)-4(t-3))/(t^2+2t-8)
=(t^2-12-t^2+4t-4t+12)/(t^2+2t-8)
=0
故K1+K2是一个常数0.

追问

第二题呢？

追答

第二题目不是已经做好了吗?

追问

好，谢谢