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A curve is represented parametrically by the equations `x=t+e^(at) and y=-t+e^(at)` when `t in R and a > 0.` If the curve touches the axis of x at the point A, then the coordinates of the point A areA. (1,0)B. `((1)/(e,0))`C. `(e,0)`D. `(2e,0)` |
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Answer» Correct Answer - D `X=t+e^(at),y=-t+e^(at)` `(dx)/(dt)=1+ae^(at),(dy)/(dx)=(-1+ae^(at))/(1+ae^(at))` At the point `A,y=0` and `(dy)/(dx)=0` for some `t=t_(1)` `becauseae^(at_(1))=1` (i). Also `0=-t+e^(at_(1))` `t_(1)=e^(at_(1))` On putting this value in Eq. (i) we get `at_(1)=1impliest_(1)=(i)/(a)` Now from eq `(i), `ae=1impliesa=(1)/(e)` Hence, `x_(A)=t_(1)+e^(at_(1))=e+e=2e` `impliesA=(2e,0)` |
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