For the following equation $\sqrt x + \sqrt y = 2$
(1) Find equation of tangent at point (a, b) on curve
Using implicit differentiation: $$y' = - \frac{√y}{√x}$$
Equation at (a, b) is: $$y - b = - \frac{\sqrt b}{\sqrt a}(x - a)$$ $$y = - x\frac{\sqrt b}{\sqrt a} + a\frac{\sqrt b}{\sqrt a} + b$$
(2) Find points where the tangent intersects at x and y axes and show the sum of these is always 4
Let y-intercept be (0, $y_0$) and x-intercept be ($x_0$, 0)
Slope = $\frac{y_0}{x_0}$
Equation of line through (a, b) is: $$y = x\frac{y_0}{x_0} – a\frac{y_0}{x_0} + b$$
Equating it to the equation of tangent: $$- x\frac{\sqrt b}{\sqrt a} + a\frac {\sqrt b}{\sqrt a} + b = x\frac{y_0}{x_0} – a\frac{y_0}{x_0} + b$$
$$\frac{\sqrt b}{\sqrt a} = - \frac {y_0}{x_0}$$
Not sure what to do after this
From $y-b=-\sqrt{\frac{b}{a}}(x-a)$ for $y=0$ we obtain $x_0=a+\sqrt{ab}$
and for $x=0$ we get $y_0=b+\sqrt{ab}.$
Id est, $$x_0+y_0=a+b+2\sqrt{ab}=(\sqrt a+\sqrt b)^2=4.$$