UPSC NDA I and II 2020 Math Solved Paper

Given parabola : y2=x.
From standard equation of parabola y2=4ax
Focus =(a,0)
Here a=

1

4

∴ Focus of parabola y2=x=(

1

4

,0)
Vertex of the parabola =(0,0)
Let P(

t2

4

,

t

2

) lies on the parabola then, its slope =tan(θ)
∴

t 2

t2 4

=tan‌θ⇒t=2‌cot‌θ
So, point P(

t2

4

,

t

2

)=(cot2θ,cot‌θ)
Its distance from vertex (0,0)
=√(cot2θ−0)2+(cot‌θ−0)2
=cot‌θ‌√cot2θ+1 =cot‌θ.cosec‌θ
=cos‌θ.cosec2θ