Question

A curve is represented parametriacally by equation $x=t+e^{at}$ & $y=-t+e^{at}$ when $t\in R$ & $a>0$. If the curve touches the axis of $x$ at the point $A$, then the coordinates of the point $A$ are

• A. (1,0)
• B. (1/e, 0)
• C. (e, 0)
• D. (2e, 0)

Answer 1

The correct option is D (2e, 0)

REF.Image.

$x=t+e^{at},y=-t+e^{at}\forall t\in R,a>0$

If curve touches the x axis at the point $A$.

Let $A$ curve is tangent at any point of x-axis

slope of tangent is parallel to x-axis.

$\frac{dx}{dt}=1+a{e}^{at},\frac{dy}{dt}=-1+a{e}^{at}$

$m_T=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-1+a{e}^{at}}{1+a{e}^{at}}$

at point $A$ slope os $0$. [$\because$ wave touches at x-axis]

So,

$m_T|A(\alpha ,0)=\frac{-1+a{e}^{at}}{1+a{e}^{at}}=0$

$\Rightarrow m_T=-1+a{e}^{at}=0$

$a{e}^{at}=1$...........(A)

Coordinates of $A$, $\lambda =\alpha$ and $y=0$

$\alpha =t+e^{at}$......(1) $0=-t+e^{at}$.........(2)

equation (1) + equation (2)

$\alpha =2e^{at}$

$e^{at}=\frac{\alpha }{2}$.........(3)

and from equation (2)

$e^{at}=t$.........(4)

from (A) and (4)

$a-t=1$............(5)

and $y=-t+e^{at}=0$ [$\because y$ coordinate is 0]

$y=-t+e^1=0$ [from (5)]

$\Rightarrow t=e$

and $x=t+e^{at}$ [given]

$x=e+e^1$

$x=2e$ and $y=0$

So, $A(2e,0)$

Answer $D$ is correct