A) \[\frac{a(1-{{t}^{2}})}{2t}\]
B) \[\frac{a({{t}^{2}}-1)}{2t}\]
C) \[\frac{a({{t}^{2}}+1)}{2t}\]
D) \[\frac{a({{t}^{2}}-1)}{t}\]
Correct Answer: B
Solution :
\[x=\frac{1-{{t}^{2}}}{1+{{t}^{2}}}\] and \[y=\frac{2at}{1+{{t}^{2}}}\] Differentiating with respect to t, we get \[\frac{dx}{dt}=\frac{(1+{{t}^{2}})(0-2t)-(1-{{t}^{2}})(0+2t)}{{{(1+{{t}^{2}})}^{2}}}=-\frac{4t}{{{(1+{{t}^{2}})}^{2}}}\] and \[\frac{dy}{dt}=\frac{(1+{{t}^{2}})2a-2at(2t)}{{{(1+{{t}^{2}})}^{2}}}=\frac{2a(1-{{t}^{2}})}{{{(1+{{t}^{2}})}^{2}}}\] Þ \[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a(1-{{t}^{2}})}{-2t};\] \[\therefore \frac{dy}{dx}=\frac{a({{t}^{2}}-1)}{2t}\].
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