Answer to Question #110662 in Calculus for Clifford Nyarko

Question

Find the sum of the x and y intercept of any tangent line to the curve √x + √y = √k

Answer 1

$$\sqrt{x}+\sqrt{y}=\sqrt{k}$$

$${d\over dx}(\sqrt{x})+{d\over dx}(\sqrt{y})={d\over dx}(\sqrt{k})$$

$${1 \over 2\sqrt{x}}+{1 \over 2\sqrt{y}}y'=0$$

$$y'=-{\sqrt{y} \over \sqrt{x}}$$

Tangent line

$$y-y_0=-{\sqrt{y_0} \over \sqrt{x_0}}(x-x_0)$$

$$y=-{\sqrt{y_0} \over \sqrt{x_0}}x+\sqrt{x_0}\sqrt{y_0}+y_0$$ $$y=-{\sqrt{y_0} \over \sqrt{x_0}}x+\sqrt{y_0}(\sqrt{x_0}+\sqrt{y_0})$$

$$y=-{\sqrt{y_0} \over \sqrt{x_0}}x+\sqrt{k}\sqrt{y_0}$$

x-intercept: $y_1=0$

$$0=-{\sqrt{y_0} \over \sqrt{x_0}}x_1+\sqrt{k}\sqrt{y_0}$$

$$x_1=\sqrt{k}\sqrt{x_0}$$

yintercept: $x_2=0$

$$y_2=-{\sqrt{y_0} \over \sqrt{x_0}}(0)+\sqrt{k}\sqrt{y_0}$$

$$y_2=\sqrt{k}\sqrt{y_0}$$

 The sum of the x and y intercept of any tangent line to the curve is

$$x_1+y_2=\sqrt{k}\sqrt{x_0}+\sqrt{k}\sqrt{y_0}=\sqrt{k}(\sqrt{x_0}+\sqrt{y_0})=$$

$$=\sqrt{k}\sqrt{k}=k$$

The sum of the x and y intercept of any tangent line to the curve is equal to $k.$

At point $(0, \sqrt{k})$ the tangent line is $x=0.$

At point $(\sqrt{k},0)$ the tangent line is $y=0.$