Question

show that the family of parabolas y2=4a(x+a) is self orthogonal. 

Joshi sir comment

y2= 4a(x+a)

so 2yy' = 4a so a = yy'/2

on putting a we get y2= 4yy'/2(x+yy'/2)      

so y2 = yy' (2x+yy')  or y = 2xy' + yy'2    (1)

now on putting -1/y' in the place of y'

we get y2 = -y/y'[2x-y/y']

so -yy'2 = 2xy' - y (2)

similarity of (1) and (2) shows that the given curve is self orthogonal 

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