Question

the shortest distance to parabola Y^2=4ax from the point (0,3)​

Answer 1

Solution :

y^2=4ax

x^2+y^2−24y+128=0

x^2+(y−12)2=16

y=tx+2at+at3

12=2at+at3

when a=1

t3+2t−12=0

(t−2)(t2+2t+6)=0

t=2

P(4,4),C(0,12)

PC=4^2+8^2= root80=4√5

c urves=4√5−r

=4(root5−1)