Problem with differentiation question

Question

The question is as follows:

Let $l$ be any tangent to the curve $\sqrt{x}+\sqrt{y} = \sqrt{k}$, where $k > 0$. Show that the sum of the $x$-intercept and the $y$-intercept of $l$ is $k$.

Answer 1

Since you already solved the problem I leave you an answer to be compared with yours.

By differentiating $\sqrt{x}+\sqrt{y} = \sqrt{k}$ we get $\frac{dx}{\sqrt{x}}+\frac{dy}{\sqrt{y}}=0$. Hence $\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$ and the tangent line at $(x,y)$ is $$Y-y=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)$$ that is $$\frac{X}{\sqrt{x}}+\frac{Y}{\sqrt{y}}=\sqrt{x}+\sqrt{y}=\sqrt{k}.$$ Now it should be easy to verify that the sum of the $x$-intercept and the $y$-intercept is $$\sqrt{k}\sqrt{x}+\sqrt{k}\sqrt{y}=\sqrt{k}\cdot \sqrt{k}=k.$$

Answer 2

WLOG, $k=1$ (otherwise, rescale). Then let $x=\cos^4t,y=\sin^4t.$

A tangent has the equation

$$\frac{x-\cos^4t}{-4\sin t\cos^3t}=\frac{y-\sin^2t}{4\cos t\sin^3t}$$ or $$\frac x{\cos^2t}+\frac y{\sin^2t}=1.$$

Answer 3

To avoid implicit differentiation.

We move $\sqrt{x}$ to the RHS and square both sides: $$\sqrt{x}+\sqrt{y}=\sqrt{k} \Rightarrow y=x-2\sqrt{xk}+k$$ Let $(x_0,y_0)$ be the tangent point. The tangent line equation is: $$\frac{y-y_0}{x-x_0}=\underbrace{1-\frac{\sqrt{k}}{\sqrt{x_0}}}_{y'(x_0)}$$ The intercepts $(\color{blue}x,\color{red}0),(\color{red}0,\color{blue}y)$ are: $$\frac{\color{red}0-y_0}{\color{blue}x-x_0}=1-\frac{\sqrt{k}}{\sqrt{x_0}}\Rightarrow \color{blue}x=x_0+\frac{-y_0\sqrt{x_0}}{\underbrace{\sqrt{x_0}-\sqrt{k}}_{-\sqrt{y_0}}}=x_0+\sqrt{x_0y_0}\\ \frac{\color{blue}y-y_0}{\color{red}0-x_0}=1-\frac{\sqrt{k}}{\sqrt{x_0}} \Rightarrow \color{blue}y=y_0+\frac{-x_0(\overbrace{\sqrt{x_0}-\sqrt{k}}^{-\sqrt{y_0}})}{\sqrt{x_0}}=y_0+\sqrt{x_0y_0}\\ \color{blue}x+\color{blue}y=x_0+y_0+2\sqrt{x_0y_0}=(\sqrt{x_0}+\sqrt{y_0})^2=k.$$