parabola-who will dare to solve

1

let the 3 vertickes be (at₁²,2at₁)

(at₂²,2at₂) , (at₃²,2at₃)

now eqtan for a tangent throught (at^2,2at) is

yt=x+at^2

substitue those 3 points to get the eqtns of tangents and then get the point of intersection

u will find dat those points satisfy the equation of the cure ve u have give i,e (3x+a)(3a+x)=y²

2

eqtn for normal is y+xt=2at+at^3

put the point (am^2,-2am) in the above eqtn and u get the normal

now u have 2 eqtns with u

eqtn of the normal and eqtn of a curve

angle between these 2 is given by the angle between the tangent at the point where the normal intersects the parabola and the normal itself

thanks wat about 3rd

3rd one

let the point be (x,y)

now find the equations of the 2 tangents that can be drawn from it to
the parabola y^2=4ax

then use the conditions given

thats all