ind the angle between the parabolas y2 4ax and x2 4by at their point of intersection other than the origin

 

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find the angle between the parabolas y2 4ax and x2 4by at their point of intersection other than the origin 

SOLUTION:

 

Let us first find the points of intersection

y2=4ax and x2=4by

Therefore

y=4bx2​. Hence (4bx2​)2=4ax

16b2x4​−4ax=0

x4−64b2ax=0

x(x3−64b2a)=0

x=0 and x=4b32​.a31​

Therefore, y=0 and y=4b31​.a32​

Hence the point of intersection other than that of origin is (x,y)=(4b32​a31​,4b31​a32​)

Now the slope of the tangent at the above point is 

2yy′=4a

y′=2y4a​

y(4b32​a31​,4b31​a32​)′​=8b31​a32​4a​

=2b3−1​a31​​

Similarly 

2x=4by′ or 

y′=2bx​

y(4b32​a31​,4b31​a32​)′​=2b4b32​a31​​=2b3−1​a31​

Hence the angle between the curves at the point of intersection

tanθ=1+m1​m2​m1​−m2​​

=1+2b3−1​a31​​.2b3−1​a31​2b3−1​a31​−2b3−1​a31​​​

=2(1+b3−2​a32​)3b3−1​a31​​

=2(b32​+a32​)3b31​a31​​

Or 

θ=tan−1⎣⎢⎡​2(b32​+a32​)

 

 

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