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, $$x_0+y_0=a+b+2\sqrt{ab}=(\sqrt a+\sqrt b)^2=4.$$


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