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Tamil Nadu Board of Secondary EducationHSC Science Class 12

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

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Question

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

Sum
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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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Meaning of Derivatives
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Chapter 7: Applications of Differential Calculus - Exercise 7.2 [Page 15]

APPEARS IN

Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 7 Applications of Differential Calculus
Exercise 7.2 | Q 9 | Page 15

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