# How to find curve C given by the equation?

Consider the curve C given by the following equation $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ where $a$ is a constant with the condition $a > 0$.

Let $(x_0, y_0)$ with be a point on C such that $x>0$ and $y>0$. Now, assume that $(x_1, 0)$ and $(0, y_1)$ be considered $x$ and $y$ such that intercepts of the tangent line to C at $(x_0, y_0)$.

My question: How to proof that $x_1 + y_1 = a$?

My try:
Currently I just found the derivative using implicit differentiation, and used the slope for its tangent line, but now I'm lost and not sure what to don.

Thanks for any suggestion.

• What have you tried? And what does "0 following" mean? Were you going to write more? – Arthur Nov 3 '17 at 20:53
• Currently I just found the derivative using implicit differentiation, and used the slope for its tangent line, but now I'm lost and not sure what to do. – okdeolxas Nov 3 '17 at 20:57
• If you've already done all that, why didn't you say so in your question? That kind of details and calculation is what separates a mediocre (or even bad) question from a good one. Also, please consider using MathJax for formatting math text. It looks a whole lot nicer. – Arthur Nov 3 '17 at 21:00
• My bad, I'm still new to this website! Thanks for the heads up – okdeolxas Nov 3 '17 at 21:07

By intrinsic differention show $dy/dx = -\sqrt{y/x}$.
The tangent line at $(x_0,y_0)$ is $y - y_0 = -\sqrt{y_0/x_0}(x - x_0)$.
Let $x=a\sin^4\theta$ and $y=a\cos^4\theta$ be a perametrization of this curve. Then $\dfrac{dy}{dx}=-\cot^2\theta.$
In order to find $x_1$ and $y_1$, first we need the equation of tangent line through $(x_0,y_0)$ which is easily followed by the point-slope form $$\dfrac{y-a\cos^4\theta_0}{x-a\sin^4\theta_0}=-\cot^2\theta_0,$$ where $\theta_0$ is the corresponding parameter. From here find $x_1$ and $y_1$ by substituting $y=0$ and $x=0$ respectively.