Question

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

Answer 1

$$\begin{gathered} \sqrt{x}+\sqrt{y}=a \\ \text{Differentiating both sides w.r.t.}x, \\ \Rightarrow \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-\sqrt{y}}{\sqrt{x}} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(\frac{a^2}{4},\frac{a^2}{4}\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(\frac{a^2}{4},\frac{a^2}{4}\right)}\text{=}\frac{-\sqrt{\frac{a^2}{4}}}{\sqrt{\frac{a^2}{4}}}\text{=-1} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{a^2}{4}=-1\left(x-\frac{a^2}{4}\right) \\ \Rightarrow y-\frac{a^2}{4}=-x+\frac{a^2}{4} \\ \Rightarrow x+y=\frac{a^2}{2} \end{gathered}$$