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Question

Prove that the two parabolas y2=4ax and x2=4by intersect (other than the origin) at an angle of tan1⎢ ⎢ ⎢ ⎢3a13b132(a23+b23)⎥ ⎥ ⎥ ⎥

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Solution

Let us first find the points of intersection
y2=4ax and x2=4by
Therefore
y=x24b. Hence (x24b)2=4ax
x416b24ax=0
x464b2ax=0
x(x364b2a)=0
x=0 and x=4b23.a13
Therefore, y=0 and y=4b13.a23
Hence the point of intersection other than that of origin is (x,y)=(4b23a13,4b13a23)
Now the slope of the tangent at the above point is
2yy=4a
y=4a2y
y(4b23a13,4b13a23)=4a8b13a23
=b13a132
Similarly
2x=4by or
y=x2b
y(4b23a13,4b13a23)=4b23a132b=2b13a13
Hence the angle between the curves at the point of intersection
tanθ=m1m21+m1m2
=2b13a13b13a1321+b13a132.2b13a13
=3b13a132(1+b23a23)
=3b13a132(b23+a23)
Or
θ=tan1⎢ ⎢ ⎢ ⎢3b13a132(b23+a23)⎥ ⎥ ⎥ ⎥

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