Let us first find the points of intersection
y2=4ax and x2=4by
Therefore
y=x24b. Hence (x24b)2=4ax
x416b2−4ax=0
x4−64b2ax=0
x(x3−64b2a)=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
=b−13a132
Similarly
2x=4by′ or
y′=x2b
y′(4b23a13,4b13a23)=4b23a132b=2b−13a13
Hence the angle between the curves at the point of intersection
tanθ=m1−m21+m1m2
=2b−13a13−b−13a1321+b−13a132.2b−13a13
=3b−13a132(1+b−23a23)
=3b13a132(b23+a23)
Or
θ=tan−1⎡⎢
⎢
⎢
⎢⎣3b13a132(b23+a23)⎤⎥
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