(a,b)在x^2=2py上,2pb=a^2
设切线方程为:y=k(x-a)+b
代人:x^2=2py得:
x^2=2pk(x-a)+2pb
x^2-2pkx+(2pka-2pb)=0
判别式△ =4p^2k^2-4(2pka-2pb)
=4p^2k^2-4(2pka-a^2)
=4(pk-a)^2
=0
pk=a
k=a/p
所以,切线方程为:y=a(x-a)/p+b
即:ax-py-a^2/2=0
(a,b)在x^2=2py上,2pb=a^2
设切线方程为:y=k(x-a)+b
代人:x^2=2py得:
x^2=2pk(x-a)+2pb
x^2-2pkx+(2pka-2pb)=0
判别式△ =4p^2k^2-4(2pka-2pb)
=4p^2k^2-4(2pka-a^2)
=4(pk-a)^2
=0
pk=a
k=a/p
所以,切线方程为:y=a(x-a)/p+b
即:ax-py-a^2/2=0