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Question
Find the angle of intersection of the curves y = 4 – x2 and y = x2.
Solution
We know that the angle of intersection of two curves is equal to the angle between the tangents drawn to the curves at their point of intersection.
The given curves are y = 4 – x2 ....(i) and y = x2 .....(ii)
Differentiating eq. (i) and (ii) with respect to x, we have
`"dy"/"dx"` = – 2x
⇒ m1 = – 2x
m1 is the slope of the tangent to the curve (i).
And `"dy"/"dx"` = 2x
⇒ m2 = 2x
m2 is the slope of the tangent to the curve (ii).
So, m1 = – 2x and m2 = 2x
Now solving equation (i) and (ii) we get
⇒ 4 – x2 = x2
⇒ 2x2 = 4
⇒ x2 = 2
⇒ x = `+- sqrt(2)`
So, m1 = – 2x
= `-2sqrt(2)` and m2 = 2x = `2sqrt(2)`
Let θ be the angle of intersection of two curves
∴ tan θ = `|("m"_2 - "m"_1)/(1 + "m"_1"m"_2)|`
= `|(2sqrt(2) + 2sqrt(2))/(1 - (2sqrt(2))(2sqrt(2)))|`
= `|(4sqrt(2))/(1 - 8)|`
= `|(4sqrt(2))/(1 - 8)|`
= `|(4sqrt(2))/(-7)|`
= `(4sqrt(2))/7`
∴ θ = `tan^-1 ((4sqrt(2))/7)`
Hence, the required angle is `tan^-1 ((4sqrt(2))/7)`.
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