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Question

If (ab)13+(ba)13=4, then the acute angle (θ) of intersection of the parabolas y2=4ax and x2=4by at a point other than the origin is

A
θ=tan138
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B
θ=tan158
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C
θ=tan137
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D
θ=tan11
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Solution

The correct option is A θ=tan138
Given parabolas,
y2=4ax and x2=4by
y416a2=4byy=4a2/3b1/3

Intersection point of the given parabolas is (0,0) and (4a1/3b2/3,4a2/3b1/3)
Slope of the tangent on the parabola y2=4ax at (4a1/3b2/3,4a2/3b1/3) is
m1=2ay=2a4a2/3b1/3=a1/32b1/3

Slope of the tangent on the parabola x2=4by at (4a1/3b2/3,4a2/3b1/3) is
m2=x2b=4a1/3b2/32b=2a1/3b1/3

So, the angle between the parabola is
tanθ=m1m21+m1m2=∣ ∣ ∣ ∣ ∣ ∣a132b132a13b131+a23b23∣ ∣ ∣ ∣ ∣ ∣=32×1(ab)13+(ba)13tanθ=32×14θ=tan138

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