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Question

Show that the two parabolas x2+4a(y2ba)=0 and y2=4b(x2a+b) intersect at right angles at a common end of the latus rectum of each.

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Solution

dydx=y

x2+4a(y2ba)=0 ...............(1)

diffrentiate the eqn.

2x+4ay=0
y=m=x2a


y2=4b(x2a+b) ..........(2)

diffrentiate the eqn.

2yy=4b

y=n=2by


solving eqn (1) & (2)
(2a,2b)


let θ be the angle at intersection point

m×n=x.2by.2a=2a.2b2b.2a=1

θ=90

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