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Question

If x=1t21+t2 and y=2at1+t2, then dydx is equal to:

A
a(1t2)2t
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B
a(t21)2t
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C
a(t2+1)2t
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D
a(t21)t
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Solution

The correct option is C a(t21)2t
Given, x=1t21+t2 and y=2at1+t2

On differentiating with respect to t, respectively, we get

dxdt=(1+t2)(02t)(1t2)(0+2t)(1+t2)2
=4t(1+t2)2

and dydt=(1+t2)2a2at(2t)(1+t2)2=2a(1t2)(1+t2)2

dydx=dydtdxdt=a(1t2)2t

dydx=a(t21)2t

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