Q2)

Equation of the tangent to y²=4ax in the parametric form

 

=>x+at²-yt=0.......(1)

 

Equation of normal to x²=4by in the parametric form can be calculated as follows

 

slope of normal at any point = -2b/x

 

parametric co-ordinates of x²=4by are 2bt,bt²

 

so, equation of normal by point slope form is

 

x+yt-2bt-bt³=0

 

since it is different parabola the parameter t would differ , so lets take it as t'

 

so, x+yt'-2bt-bt' ³=0........(2)

 

comparing the coeff of x,y and the const terms of equations 1 and 2 we get.

 

2bt+bt³=at²

 

bt²-at+2b=0

 

since t is real

 

a²>=8b²