TS EAMCET 11-Sep-2020 Shift 1 Solved Paper

Given curve is
x=a(θ+sin‌θ),y=a(1−cos‌θ)
∴‌‌‌

dy

dx

=‌

sin‌θ

1+cos‌θ

=tan‌

θ

2

∴ Slope of tangent and normal drawn to the given curve at P(θ=‌

π

2

) is mT=1 and mN=−1 respectively.
Now, equation of tangent to the given curve at point P(a(‌

π

2

+1),a) is y−a=1[x−a(

π

2

+1)]
∴ Point A is (‌

aπ

2

,0)
and equation of normal to the given curve at point P‌ is ‌y−a=−1[x−a(‌

π

2

+1)]
So, point B is (a(‌

π

2

+2),0)
Therefore area of =‌

1

2

|a[‌

aπ

2

−‌

aπ

2

−2a]|=a2