Implicit differentiation q

Question

Show that the sum of the x and y intercepts of any tangent line to the curve $x^{1/2} + y^{1/2} = 4$ is equal to 16

So far I have found the derivative, $-\frac{\sqrt{y}}{ \sqrt{x}}$, but am having trouble as to how I would found the point of tangency as as that is what I would think you do next?

Answer 1

$x^{\frac12}+y^{\frac12}=4 \Rightarrow \frac{dy}{dx}=-\sqrt{\frac{y}{x}}$.

Now, the equation of a straight line that passes through a point $(x_1,y_1)$ on the curve is given by $$y-y_1=-\sqrt{\frac{y_1}{x_1}}(x-x_1)$$

The $y$ intercept, $y_0$ occurs at $x=0$. Thus,

$$\begin{align} y_0-y_1 & = -\sqrt{\frac{y_1}{x_1}}(0-x_1) \Rightarrow y_0 = y_1+\sqrt{x_1y_1} \end{align}$$

Similarly, the $x$ intercept occurs at $y=0$. Thus,

$$\begin{align} 0-y_1 & = -\sqrt{\frac{y_1}{x_1}}(x_0-x_1) \Rightarrow x_0 = x_1+\sqrt{x_1y_1} \end{align}$$

Adding the intercepts, we find

$$\begin{align} x_0+y_0 & =x_1+2\sqrt{x_1y_1}+y_1 \\ &=(\sqrt{x_1}+\sqrt{y_1})^2 \\ & = (4)^2 \\ & = 16 \end{align}$$

as was to be shown!

Answer 2

$\left(x^{\frac{1}{2}} + y^{\frac{1}{2}}\right)' = 0 \Rightarrow \dfrac{1}{2\sqrt{x}}+\dfrac{y'}{2\sqrt{y}} = 0 \Rightarrow y' = - \dfrac{\sqrt{y}}{\sqrt{x}} \Rightarrow y - y_1 = m(x-x_1) \Rightarrow y - y_1 = -\dfrac{\sqrt{y_1}}{\sqrt{x_1}}\left(x - x_1\right)$. Put $x = 0 \Rightarrow y_{\text{int}} = y_1 + \sqrt{x_1y_1} = \sqrt{y_1}\left(\sqrt{y_1}+\sqrt{x_1}\right) = 4\sqrt{y_1}$. Put $y = 0 \Rightarrow x_{\text{int}} = \sqrt{x_1}\left(\sqrt{x_1}+\sqrt{y_1}\right) = 4\sqrt{x_1}$. Thus: $x_{\text{int}} + y_{\text{int}} = 4\left(\sqrt{x_1}+\sqrt{y_1}\right) = 4\cdot 4 = 16$

Answer 3

differencing $$x^{1/2} + y^{1/2} = 4 \to x^{-1/2}\, dx + y^{-1/2}\, dy = 0$$ the tangent at $(a, b)$ such that $$\sqrt a + \sqrt b = 4 $$ is given by $$ 0=\frac{x-a}{\sqrt a} + \frac{y - b }{\sqrt b}= \frac{x}{\sqrt a} + \frac{y}{\sqrt b} - \sqrt a - \sqrt b=\frac{x}{\sqrt a} + \frac{y}{\sqrt b} - 4$$

so the $x$-intercepts is $(4\sqrt a, 0)$ and $y$-intercepts is $(0, 4\sqrt a)$ therefor the sum of the intercepts is $$4(\sqrt a + \sqrt b) = 4 \times 4 = 16. $$