Question

A curve is represented parametrically by $x=t+e^{at}$ and $y=-t+e^{at},t\in R$ and $a>0$. Then, the curve touches the $x-$ axis at

• A. $(1,0)$
• B. $(\frac{1}{e},0)$
• C. $(e,0)$
• D. $(2e,0)$

Answer 1

The correct option is D $(2e,0)$

Given curve is $x=t+e^{at};y=-t+e^{at}$
Let the curve touches $x-$ axis at $(x_1,0)$, then the slope of the tangent at this point is $0$,
Now,
$\frac{dx}{dt}=1+a{e}^{at}\frac{dy}{dt}=-1+a{e}^{at}\Rightarrow \frac{dy}{dx}=\frac{-1+a{e}^{at}}{1+a{e}^{at}}$
At $(x_1,0)$,
$\frac{-1+a{e}^{a{t}_1}}{1+a{e}^{a{t}_1}}=0\Rightarrow a{e}^{a{t}_1}=1\cdots \left(1\right)$
Also,
$y=0\Rightarrow -t_1+e^{a{t}_1}=0\Rightarrow e^{a{t}_1}=t_1$
Using equation $\left(1\right)$, we get
$\Rightarrow t_1=\frac{1}{a}$
Again using equation $\left(1\right)$, we get
$ae=1\Rightarrow a=\frac{1}{e}\Rightarrow t_1=e\Rightarrow x_1=t_1+e^{a{t}_1}\therefore x_1=e+e=2e$
Hence, the point of contact is $(2e,0)$