Question

Find the equation of the tangent to the curve $\sqrt{x}+\sqrt{y}=a$ at the point $(\frac{a^2}{4},\frac{a^2}{4})$.

Answer 1

$\sqrt{x}+\sqrt{y}=a$

Differentiate both sides w.r.t $x,$

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

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

$(x_1,y_1)=(\frac{a^2}{4},\frac{a^2}{4})$

Slop of tangent , $m={\left(\frac{dy}{dx}\right)}_{(\frac{a^2}{4},\frac{a^2}{4})}=\frac{-\sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}=-1$

Equation of tangent,

$y-y_1=m(x-x_1)$

$\Rightarrow$ $y-\frac{a^2}{4}=-1(x-\frac{a^2}{4})$

$\Rightarrow$ $y-\frac{a^2}{4}=-x+\frac{a^2}{4}$

$\Rightarrow$ $x+y=\frac{a^2}{2}$