Find the angle to intersection of the following curves :
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
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θ = tan – 1()