Question

The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are_______________.

Answer 1

Let (h, k) be the point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3.

y = x log_ex

Differentiating both sides with respect to x, we get

$\frac{dy}{dx}=x\times \frac{1}{x}+{\log }_{e}x\times 1$

$\Rightarrow \frac{dy}{dx}=1+{\log }_{e}x$

It is given that, the normal is parallel to the line 2x − 2y = 3.

∴ Slope of normal at (h, k) = Slope of the given line

$\Rightarrow -\frac{1}{{\left({\displaystyle \frac{dy}{dx}}\right)}_{\left(h,k\right)}}=-\frac{2}{\left(-2\right)}$

$\Rightarrow -\frac{1}{1+{\log }_{e}h}=1$

$\Rightarrow 1+{\log }_{e}h=-1$

$\Rightarrow {\log }_{e}h=-2$

$\Rightarrow h=e^{-2}=\frac{1}{e^2}$

Now, (h, k) lies on the given curve.

$\therefore k=h{\log }_{e}h$ .....(1)

Putting $h=\frac{1}{e^2}$ in (1), we get

$k=\frac{1}{e^2}\times {\log }_e\frac{1}{e^2}$

$\Rightarrow k=\frac{1}{e^2}\times {\log }_e{e}^{-2}$

$\Rightarrow k=-\frac{2}{e^2}\left(\log a^b=b\log a\right)$

So, the coordinates of the required points are $\left(\frac{1}{e^2},-\frac{2}{e^2}\right)$.

Thus, the coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $\left(\frac{1}{e^2},-\frac{2}{e^2}\right)$.


The coordinates of a point on the curve y = x log_ex at which the normal is parallel to the line 2x − 2y = 3 are $\underline{\left(\frac{1}{e^2},-\frac{2}{e^2}\right)}$.