A beautiful Parabolaaas

As, $y^2$ =4ax, co-ordinates of points on latus rectum are ‘P’ (a,2a) and ‘Q’(a,-2a) .

Equation of lines passing through (a,2a) and ‘O’ origin is y=2x.

Equation of lines passing through (a,-2a) and origin is y=-2x.

So they are ⊥ to each other. Hence area of equilateral triangle △POQ will be

1/2{ $1^2*a^2+4*a^2\$ } =1/2{(5 $a^2$ )}

Now normal drawn to points ‘P’ and ‘Q’ have equations x+y=3a and x-y=3a respectively.

So they intersect at point (3a,0). Now a parabola is drawn whose vertex is at (3a,0) and focus is (4a,0) .

Also it is in the direction of previous one. So equation of this parabola is $y^2$ =4a(x-3a).

So points of latus rectum on this parabola are ‘A’(4a,2a) and ‘B’(4a,-2a). so area of equilateral triangle

△ AOB is 1/2{ $4^2*a^2+4*a^2\$ }= $10a^2$ .

Thus for ‘t’th parabola , area of triangle will be 1/2{ $n^2*a^2+4*a^2\$ }

Where , n=1+(t-1)(4-1).

Because x-co-ordinate will increase as 1,4,7,…..

So, n=1+3t-3=3t-2 ........................................... eq:-1

For area to be 1156 times

1/2{ $n^2*a^2+4*a^2\$ }=1156 1/2(5 $a^2$ )

This equals to , $n^2$ +4=5780, therefore , n=76

Therefore , putting in eqn 1

We get, t=26