Question
The sum of the intercepts of a tangent to $\sqrt{x}+\sqrt{y}$ $=\sqrt{a}, a>0$ upon the coordinate axes is(a) $2 a$(b) $a$(c) $a / 2$(d) $\sqrt{a}$
Step 1
We get \[\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0\] From here we get $\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}$. Show more…
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