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Question

Question: Find the angle of intersection of the following curves.\[\ y^{2} = x\] and \[x^{2} = y\]....

Find the angle of intersection of the following curves. y2=x\ y^{2} = x and x2=yx^{2} = y.

Explanation

Solution

In this question, we need to find the angle of intersection of the two given curves. In order to find the angle of intersection, first we need to find the point of intersection of the two given curves. Then we need to find the slope of both the curves. By finding this, we can find the angle of intersection of the curves.

Complete step-by-step solution:
Given ,
y2=xy^{2} = x ……………………………….(1)
x2=yx^{2} = y ……………………………… (2)
By substituting (2) in (1),
We get,
(x2)2=x\left( x^{2} \right)^{2} = x
x4=xx^{4} = x
By bringing all the terms to one side,
We get,
x4x=0x^{4} – x = 0
Taking xx as common,
x(x31)=0x\left( x^{3} – 1 \right) = 0
Thus we get the value of xx as
 (x=0) or (x31=0)\ (x = 0) \ or\ (x^{3} – 1 = 0)
By simplifying,
We get,
 (x=0) or (x3=1)\ (x = 0)\ or\ (x^{3} = 1)
Thus we get the values of xx are 00 and 11
Substituting the values of xx in (2)
We get,
y=0y = 0 or y=1y = 1
Therefore the point of intersection are (0, 0)(0,\ 0) and (1, 1)\ (1,\ 1)
Then we have differentiate the given curves,
From (1),
y2=xy^{2} = x
Differentiating (1) w.r.t to x
2y(dydx)=12y\left( \dfrac{{dy}}{{dx}} \right) = 1
dydx=12y\dfrac{{dy}}{{dx}} = \dfrac{1}{2y}
Differentiating (2) w.r.t to x
2x=dydx2x = \dfrac{{dy}}{{dx}}
At (0, 0) (0,\ 0)\ , the slope of the tangent to the curve y2=xy^{2} = x is parallel to the yaxisy – axis and the tangent to the curve x2=yx^{2} = y is parallel to xaxisx – axis.
At (1, 1)(1,\ 1), slope of the tangent to the curve y2 = xy^{2}\ = \ x is equal to 12\dfrac{1}{2} and that of x2 = yx^{2}\ = \ y is 22.
Thus we get,
m1=12m{_1}= \dfrac{1}{2}
m2=2m{_2}= 2
If θ\theta is the angle of intersection of the curves,
tanθ= (m1m2)(1+m1m2)tan\theta = \ |\dfrac{(m{_1}m{_2})}{(1 + m{_1}m{_2})}|
By substituting the values,
We get,
tanθ=(1)2tan\theta = \left| \dfrac{\left( 1\right)}{2} \right|
tanθ=(12)tan\theta = \left| \left( \dfrac{1}{2} \right)\right|
Thus
θ=tan1(12)\theta = \tan^{- 1}\left( \dfrac{1}{2} \right)
Final answer :
The angle of intersection is tan1(12)\tan^{- 1}\left(\dfrac{1}{2} \right)

Note: The concept used to find the angle of intersection is tangents and the normal. The angle of intersection is nothing but the angle of the line formed with the horizontal. The formula for the angle of intersection is tanθ=m\tan \theta = m