Question

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

Answer 1

Let the required point be $(x_1,y_1)$

Given equation of curve is $\sqrt{x}+\sqrt{y}=4$

Since $(x_1,y_1)$ lies on the curve we get,

$\Rightarrow \sqrt{x_1}+\sqrt{y_1}=4$ ...$\left(1\right)$

Consider the curve $\sqrt{x}+\sqrt{y}=4$

On differentiating w.r.t x we get

$\Rightarrow \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$

$\Rightarrow \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$

Given that tangent is equally inclined to the axes

$\theta ={45}^0\Rightarrow tan\theta =1$

$\Rightarrow (\frac{dy}{dx}{)}_{(x_1,y_1)}=1$

$\Rightarrow -\sqrt{\frac{y_1}{x_1}}=1$

Squaring on both sides, we get

$\Rightarrow x_1=y_1$

From $\left(1\right)$ we have

$\Rightarrow \sqrt{x_1}+\sqrt{y_1}=4$

$\Rightarrow \sqrt{x_1}+\sqrt{x_1}=4$

$\Rightarrow 2\sqrt{x_1}=4$

$\Rightarrow \sqrt{x_1}=2$

$\Rightarrow \sqrt{x_1}=2$

$\Rightarrow x_1=y_1=4$

Hence the required point is $(4,4)$