The correct option is C 3
Given equations are
x2+y2=5.....(i)
and y2=4x.....(ii)
On solving Eqs. (i) and (ii), we get
x=−5,1
at x=−5,y2=−20 (imaginary value)
∴ at x=1,y2=4
⇒y=±2
Hence, point of intersection are (1,2) and (1,−2)
On differentiating Eq. (i) w.r.t. x, we get
2x+2ydydx=0
⇒dydx=−xy
∴m1=(dydx)(1,2)=−12
And on differentiating Eq. (ii) w.r.t. x, we get
2ydydx=4
⇒dydx=2y,m2=(dydx)(1,2)=22=1
Now, tanθ=∣∣∣m1−m21+m1m2∣∣∣
=∣∣
∣
∣∣−12−11−12∣∣
∣
∣∣=∣∣
∣
∣∣−3212∣∣
∣
∣∣=3