Find the angle to intersection of the following curves :
and x2 + y2 = ab
Given:
Curves 
+ 
1 ...(1)
& x2 + y2 = ab ...(2)
Second curve is x2 + y2 = ab
y2 = ab – x2
Substituting this in equation (1),
+ 
1
1
x2b2 + a3b – a2x2 = a2b2
x2b2 – a2x2 = a2b2 – a3b
x2(b2 – a2) = a2b(b – a)
x2
x2
x2
∴a2 – b2 = (a + b)(a – b)
x
...(3)
since , y2 = ab – x2
y2 = ab – (
)
y2
y2
y = ±
...(4)
since ,curves are 
+ 
1 & x2 + y2 = ab
Differentiating above w.r.t x,
⇒ 
.
= 0
⇒ 
.
= 
⇒ 
⇒ 
⇒ m1
...(5)
Second curve is x2 + y2 = ab
⇒ 2x + 2y.
0
⇒ m2
...(6)
Substituting (3) in (4), above values for m1 & m2,we get,
At (
, 
) in equation(5),we get
⇒ m1
At (
, 
) in equation(6),we get
m2
when m1
& m2
tanθ
tanθ
tanθ
tanθ
tanθ
tanθ
tanθ
θ = tan – 1(
)
1