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Question

If x+y=t1t,x2+y2=t2+1t2, then dydx is equal to

A
yx
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B
1x
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C
1x2
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D
1x2
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Solution

The correct option is A yx

We have,

x+y=t1t......(1)

x2+y2=t2+1t2.......(2)

By equation (2) to,

x2+y2=t2+1t22+2

x2+y2=(t1t)2+2

x2+y22=(t1t)2.......(3)

By equation (1) and (3) to, and we get,

x2+y22=(x+y)2

x2+y22=x2+y2+2xy

2=2xy

xy=1

On differentiating and we get,

xdydx+ydxdx=0

xdydx+y(1)=0

xdydx=y

dydx=yx

Hence, this is the answer.

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