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Question

The orthogonal trajectory of y2=4ax, (a being the parameter) is:
(where c is an arbitrary constant)

A
2x2+3y2=c
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B
2x2+y2=c
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C
2x2y2=c
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D
x2+y2=c
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Solution

The correct option is B 2x2+y2=c
y2=4ax(i)2ydydx=4a(ii)
Eliminating a from equation (i) and (ii), we have
y2=2ydydxxy=2xdydx
To find the orthogonal trajectory, replacing dydx by dxdy, we get
y=2(dxdy)x
2x dx+y dy=0
Integrating both sides, we get
x2+y22=C2x2+y2=c, (let c=2C)

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