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

‚Äčy=qe-xpdydx=-qpe-xp

When the curve crosses the y-axis, x=0.
At x=0, y=qe-0p, i.e., y=q

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

 At (0, q), dydx=-qpe0=-qp

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

y-q=-qpx-0xp+yq=1

Hence Proved.

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