Question

求k为何值时,直线y=kx+3与椭圆x^2/9+y^2/4=1,1)相交于两点

Answer 1

y=kx+3

x^2/9+y^2/4=1

x^2/9+(kx+3)^2/4=1

(1/9+k^2/4)*x^2+3k/2x+5/4=0,该方程有两个不同的根,所以判别式大于0,既有：

9k^2/4-4(1/9+k^2/4)*5/4>0

9k^2>5

k^2>5/9

k>5^0.5/3或 k