X int and Y int add to C Problem

Question

Let $c>0$ and let $L$ be tangent to the curve $\sqrt{x}+\sqrt{y}=\sqrt{c}$. Show that the sum of the x-int of $L$ and the y-int of $L$ is equal to $c$

Solution

First get the tangent line of L. Get the derivative first.

• $0.5(x^{-0.5})+0.5(y^{-0.5})\frac{dy}{dx}=0$
• $\frac{dy}{dx}=\frac{-0.5(x^{-0.5})}{0.5(y^{-0.5})}$
• $\frac{dy}{dx}=\frac{-(\frac{1}{\sqrt{x}})}{\frac{1}{\sqrt{y}}}$
• $\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}$

Then substitute x and y as $a,b$ so $\sqrt{a}+\sqrt{b}=\sqrt{c}$

Tangent line $y-b=-\frac{\sqrt{b}}{\sqrt{a}}(x-a)$ So now. to get the x-int, take $y=0$

• $0-b=-\frac{\sqrt{b}}{\sqrt{a}}(x-a)$
• $x=a+\sqrt{ab}$ For the y-int, take $x=0$
• $y-b=-\frac{\sqrt{b}}{\sqrt{a}}(0-a)$
• $y=b+\sqrt{ab}$ Then $x+y=a+\sqrt{ab}+b+\sqrt{ab}$
• $=a+b+2\sqrt{ab}$
• $=(\sqrt{a}+\sqrt{b}{)}^2$
• $=(\sqrt{c}{)}^2$
• $=c$ $\square$