BITSAT 2006 Solved Question Paper

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+2y

dy

dx

=0

dy

dx

=−

x

y

∴m1=(

dy

dx

)(1,2)=−

1

2

And on differentiating Eq. (ii) w.r.t. x, we get
2y

dy

dx

=4
⇒

dy

dx

=

2

y

,m2=(

dy

dx

)(1,2)=

2

2

=1
Now, tan‌θ=|

m1−m2

1+m1m2

|
=|

− 1 2 −1

1− 1 2

|=|

− 3 2

12

|=3