Question

The area of the triangle formed by the tangent, normal at $P(1,1)$ on the curve $\sqrt{x}+\sqrt{y}=2$ with the $x$ -axis is

Answer 1

Given, $\sqrt{x}+\sqrt{y}=2$

on differentiating w.r.t $x$
$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0$
$\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$
Slope of tangent at given point $=-1$
$\Delta$Area$=\frac{1}{2}\times$ length of Tangent$\times$length of Normal

$=\frac{1}{2}\left|1\sqrt{1+\frac{1}{(-1{)}^2}}\right|\left|1\sqrt{1+(-1{)}^2}\right|=1$ sq. units.