题错了吧
x2/3+y2=4,4应该是1吧?
如果是如下解
设p(x0,y0)设过这一点的直线为y-y0=k(x-x0)
y=kx-kx0+y0,设b=-kx0+y0
y=kx+b
椭圆方程同乘3得x²+3y²-3=0
把y=kx+b代入得
x²+3k²x²+6kbx+3b²-3=0
△＝0,△除以4得
9k²b²-3b²-9k²b²+9k²+3=0
9k²-3b²+3=0
3k²-b²+1=0
b=-kx0+y0
3k²-k²x0²+2kx0y0-y0²+1
△=3y0²+x0²-3＞0
两根k1·k2=(-y0²+1)/(3-x0²)
x0²+y0²=4,y0²=4-x0²
k1·k2=(x0²-3)/(3-x0²)=-1
随意两直线垂直
m⊥n