Question: $P(x_1,y_1)$ is a point on the curve $y=e^{ – x}$. The tangent to the curve at P passes through the origin. Find the coordinates of $P$. NB: I am assuming the question is asking for an integer or some sort of value for the coordinates.
My working out: Firstly I tried the first derivative, which is $y=-e^{ – x}$ and the subbing in $x=x_1$ from point $P$ – this is this gradient for point $P$
but that just goes to get: $y=-e^{ – x_1}$, and if that's it then subbing it into the point-gradient formula ($y-y_1=m(x-x_1)$)
leading to:
$y-y_1=y=-e^{ – x_1}(x-x_1)$ etc… BUT continuing on does not get me the coordinates of $P$.
or I tried the simple method which was; $m=\frac{y_2-y_1}{x_2-x_1}$ by using the point $P$ coordinates and the origion (assuming it's $O(0,1)$) from the original formula. But I have no idea how I would go on from here.
Best Answer
You are very close we have that $y_1=e^{-x_1}$. Then we have the tangent line at a point is $$ y-e^{-x_1}=-e^{-x_1}(x-x_1) $$ But now the tangent line has to go through the origin so we get, $$ x_1e^{-x_1}=-e^{-x_1}. $$ This means $x_1 =-1$ so we get the point $(-1,e^{1})$