Question

求过点P（2,-1）,且与椭圆x平方/6+y平方/3=1相切的直线方程

Answer 1

x^2/6+y^2/3=1

P直线y=k(x-2)-1

3x^2+6[k(x-2)-1]^2=18

(3+6k^2)x^2+6[k^2(x^2-4x+4)-2k(x-2)+1]-18=0

(3+6k^2)x^2+(-24k^2-12k)x+(24k^2+24k-12)=0

判别式

(-24k^2-12k)^2-4*(3+6k^2)(24k^2+24k-12)=0

(6k^2+3k)^2-(3+6k^2)(6k^2+6k-3)=0

(2k^2+k)^2-(1+2k^2)(2k^2+2k-1)=0

k^2-2k^2-2k+1+2k^2=0

k^2-2k+1=0

k=1

切线：y=(x-2)-1,y=x-3