# show that the line x/p+ y/q =1, touches the curve y=e^(-x/p) at the point where it crosses the y axis

$y=q{e}^{-\frac{x}{p}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=-\frac{q}{p}{e}^{-\frac{x}{p}}$

When the curve crosses the y-axis, x=0.
At x=0,

So, the curve crosses the y-axis at (0, q).

Now, equation of the tangent to y at (0, q) is:-

$y-q=-\frac{q}{p}\left(x-0\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{p}+\frac{y}{q}=1$

Hence Proved.

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