Question

Find the angle between the rectangular hyperbola xy = 2 and the parabola x² + 4y = 0

Answer 1

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)`