Question

Find the coordinates of the point on the curve $\sqrt{x}+\sqrt{y}=4$ at which tangent is equally inclined to the axes.

Answer 1

We have$\sqrt{x}+\sqrt{y}=4\Rightarrow x^{\frac{1}{2}}+y^{\frac{1}{2}}=4\Rightarrow \frac{1}{2}.\frac{1}{x^{\frac{1}{2}}}+\frac{1}{2}.\frac{1}{Y^{\frac{1}{2}}}.\frac{dy}{dx}=0\therefore \frac{dy}{dx}=-\frac{1}{2}.x^{-\frac{1}{2}}2.y^{\frac{1}{2}}=-\sqrt{\frac{y}{x}}$
Since, tangent is equally inclined to the axes.
$\therefore \frac{dy}{dx}=\pm 1\Rightarrow -\sqrt{\frac{y}{x}}=\pm 1\Rightarrow \frac{y}{x}=1\Rightarrow y=xFromEq.\left(i\right).,\sqrt{y}+\sqrt{y}=4\Rightarrow 2\sqrt{y}=4\Rightarrow 4y=16\therefore y=4andx=4$
So, the required coordinates are (4, 4).