Question

高数，这道题怎么求，求个过程

Answer 1

设直线标准式和椭圆方程联立，整理出一个方，程，然后△等于0，注意一下二次项的系数，可能会有讨论。接下来应该会做了。哦对了直线还过一个点，可以更简单一下

Answer 2

x^2/6 +y^2/3 =1 (1)
切线方程, 过点(4,-1)
y+1=m(x-4) (2)
sub (2) into (1)
x^2/6 +y^2/3 =1
x^2/6 + (mx-4m-1)^2/3 =1
x^2 + 2(mx-4m-1)^2 =6
(1+2m^2)x^2 -4m(4m+1)x + 2(4m+1)^2 - 6 =0
△=0

16m^2.(4m+1)^2 - 4(1+2m^2)[2(4m+1)^2 - 6] =0
(4m+1)^2 . [ 16m^2 -8(1+2m^2) ] - 24(1+2m^2)=0
-8(4m+1)^2 - 24(1+2m^2)=0
(4m+1)^2 - 3(1+2m^2)=0
16m^2+8m+1 -3-6m^2=0
10m^2+8m-2 =0
5m^2+4m-1 =0
(5m-1)(m+1)=0
m=1/5 or -1
ie
切线方程, 过点(4,-1)
y+1=(1/5)(x-4) or y+1=-(x-4)