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The curve with equation $y = e^{-ax}(\tan x)$, where $a$ is a positive constant, has only one point in the interval $0 < x < \pi/2$ at which the tangent is parallel to the x-axis. Find the value of a and state the exact value of the $x$-coordinate at this point.

I have differentiated using the product rule $\tan^2 x - a\tan x + 1 = 0$. But I am not sure how to proceed. I've also got $y= \frac{2}{\sin 2x}$ but I'm also stuck at this step.

Please send help. SOS

user487026
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Let $t=\tan x$, $t\in(0,+\infty)$. Need to prove the solution of $t^2-at+1=(t-{a\over 2})^2+1-{a^2\over 4}=0$ has only one point. A quadratic equation has only one point if and only if $1-{a^2\over 4}=0$, hence $a=2$, then $t=1\Rightarrow \tan x=1\Rightarrow x={\pi\over 4}$.

yahoo
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