AP EAMCET Engineering 2012 Solved Paper

Given equation of circle isx2+y2=4
On differentiating w.r.t. x, we get
2x+2y

dy

dx

=0
⇒

dy

dx

=−

x

y

⇒ (

dy

dx

)(1,√3)=−

1

√3

∴ Equation of tangent at (1,√3) is
y−√3=−

1

√3

(x−1)
⇒ √3y−3=−x+1
⇒ x+√3y=4 .......(i)
and equation of normal at (1,√3) is
y−√3=√3(x−1)
⇒ √3x−y=0 .......(ii)
∴ Intersection point of Eqs. (i) and (ii) is (1,√3).
∴ Area of Delta‌O‌A‌B=

1

2

×OB×AD=

1

2

×4×√3
=2√3