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Question

Find the angle between the prabolas y2=4ax and x2=4by at their point of intersection other than origin.


Solution

The equations of two parabolas are y2=4ax and x2=4by.
Now, x2=4byy=x24b
Substituting this value of y in y2=4ax, we get
(x24b)2=4ax
x464ab2x=0
x(x364ab2)=0
x=0 or x3=64ab2
x=0 or x=4a1/3b2/3
So, y=x24b gives y=0,y=4a2/3b1/3.

Thus, the two curves intersect at point P(4a1/3b2/3,4a2/3b1/3) other than origin O(0,0).
Now, y2=4ax and x2=4by
Differentiate above equations w.r.t. x,
2ydydx=4a and 2x=4bdydx
dydx=2ay and dydx=x2b

m1=dydx=2ay=2a4a2/3b1/3=12(ab)1/3  and 
m2=dydx=x2b=4a1/3b2/32b=2(ab)1/3

Let θ be the angle between the tangents to the parabolas y2=4ax and x2=4by at P. Then,
tanθ=m1m21+m1m2
        =∣ ∣ ∣ ∣ ∣12(ab)1/32(ab)1/31+12(ab)1/3×2(ab)1/3∣ ∣ ∣ ∣ ∣ 
        =∣ ∣ ∣ ∣ ∣32(ab)1/31+(ab)2/3∣ ∣ ∣ ∣ ∣
tanθ=3a1/3b1/32(a2/3+b2/3)
θ=tan1(3(ab)1/32(a2/3+b2/3))

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