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Question
Find the angle between the rectangular hyperbola xy = 2 and the parabola x2 + 4y = 0
Solution
Given curves are xy = 2 ........(1)
x2 + 4y = 0 ........(2)
Now solving (1) and (2)
Substituting (1) in (2)
⇒ x2 + `4(2/x)` = 0
x3 + 8 = 0
x3 = – 8
x = – 2
Substituting in (1)
⇒ y = `2/(-2)` = – 1
∴ Point of intersection of (1) and (2) is (– 2, – 1)
xy = 2
⇒ y = `2/x` ........(1)
Differentiating w.r.t. ‘x’
`("d"y)/("d"x) = - 2/x^2`
Slope of the tangent 'm1' = `(("d"y)/("d"x))_(((-2, -1)))`
= `- 2/4 = - 1/2`
x2 + 4y = 0
⇒ y = `- x^2/4`
Differentiating w.r.t. 'x'
`("d"y)/("d"x) = - (2x)/4 = - x/2`
Slope of the tangent 'm2' = `(("d"y)/("d"x))_(((-2, -1)))`
= `2/2`
= 1
The angle between the curves
θ = `tan^-1 |("m"_1 - "m"_2)/(1 + "m"_1"m"_2)|`
θ = `tan^-1|((-1)/2 - 1)/(1 - 1/2)|`
`tan^-1|(- 3/2)/(1/2)|`
θ = `tan^1 (3)`
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