圆锥曲线问题抛物线y^2=2px(p>0)上一点M(x0,y0)处切线方程是?
2019-05-27
圆锥曲线问题
抛物线y^2=2px(p>0)上一点M(x0,y0)处切线方程是?
优质解答
设点斜式k(y-y0)=x-x0
x=x0+k(y-y0)
代入y^2=2px
y^2=2p[x0+k(y-y0)]
y^2-2pky-2p(x0-ky0)=0
Δ=0
4(pk)^2+8p(x0-ky0)=0
4(pk)^2-8pky0+8px0=0
[x0=y0^2/2p]
4(pk)^2-8pky0+4y0^2=0
(pk-y0)^2=0
k=y0/p
y0/p(y-y0)=x-x0
yy0=p(x-x0)+y0^2
yy0=p(x-x0)+2px0
yy0=p(x+x0)
设点斜式k(y-y0)=x-x0
x=x0+k(y-y0)
代入y^2=2px
y^2=2p[x0+k(y-y0)]
y^2-2pky-2p(x0-ky0)=0
Δ=0
4(pk)^2+8p(x0-ky0)=0
4(pk)^2-8pky0+8px0=0
[x0=y0^2/2p]
4(pk)^2-8pky0+4y0^2=0
(pk-y0)^2=0
k=y0/p
y0/p(y-y0)=x-x0
yy0=p(x-x0)+y0^2
yy0=p(x-x0)+2px0
yy0=p(x+x0)