JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given curves

x2

a2

+

y2

b2

=1 ...(i)
and x2+y2=ab ...(ii)
From Eqs. (ii)
y2=ab−x2
From Eq. (i),
b2x2+a2(ab−x2)=a2b2
(b2−a2)x2=a2b(b−a)
⇒x2=

a2b

a+b

y2=ab−

a2b

a+b

=

ab2

a+b

Point of intersection (√

a2b

a+b

,√

ab2

a+b

)
Now, differentiating Eq. (i) w.r.t. x, we have

dy

dx

=−

b2x

a2y

=m1 (Let)
and differentiating Eq. (ii) w.r.t. x,

dy

dx

=

−x

y

=m2 (Let)
Let angle be θ, Then, =|

a−b

√ab

|
⇒θ=tan−1(

a−b

√ab

)