# Analytic geometry and calculus question

The tangent to the curve $x^{0.5} + y^{0.5} = a^{0.5}$ ,$(a>0)$ intersects the $x$ axis in point $M_1$ , and the $y$ axis in point $M_2$. Prove that for any point on the curve $|OM_1|+|OM_2| = a$ , where $O$ is the point of the beginning of axis, and $|OM_1| ,|OM_2|$ lengths of the segments. Is there a particular method for this kind of proof? I did found the tangent equation but it didn't help much.

• If you have the equation of the tangent line for a point (X, Y) on this curve , you can find the $\ x-$ and $\ y-$ intercepts. That will tell you the lengths of $\ OM_1 \$ and $\ OM_2 \$ . – colormegone Mar 15 '14 at 14:52

Implicit differentiation of the curve equation yields: $$\frac1{2\sqrt{x}}+\frac1{2\sqrt{y}}\times y^\prime=0 \\ y^\prime =−\frac{\frac1{2\sqrt{x}}}{\frac1{2\sqrt{y}}}=-\frac{\sqrt{y}}{\sqrt{x}}$$
So for the tangent to the curve at a point $(x_0, y_0)$ on the curve, we get:
$$y−y_0= -\frac{\sqrt{y_0}}{\sqrt{x_0}} (x−x_0 ) \\ \frac{y}{\sqrt{y_0}}−\sqrt{y_0 }=−\frac{x}{\sqrt{x_0}}+\sqrt{x_0} \\ \frac{x}{\sqrt{x_0}}+\frac{y}{\sqrt{y_0}}=\sqrt{x_0}+\sqrt{y_0}=\sqrt{a}$$ Now, to find $M_1$ and $M_2$ we substitute $y=0$ and $x=0$ respectively.
With $y=0$: $$\frac{x}{\sqrt{x_0} }=\sqrt{a} \\ x=\sqrt{ax_0} \\ M_1 (\sqrt{ax_0},0)$$ With $x=0$: $$\frac{y}{\sqrt{y_0 }}=\sqrt{a} \\ y=\sqrt{ay_0} \\ M_2 (0,\sqrt{ay_0}) \\ \therefore \left|OM_1 \right|+\left|OM_2 \right|=\sqrt{ax_0 }+\sqrt{ay_0 }=\sqrt{a} (\sqrt{x_0 }+\sqrt{y_0 })=\sqrt{a} \times \sqrt{a}=a$$