The tangent to the curve y = (sqrt(a)- sqrt(x))², a > 0

The tangent to the curve $y = \bigl(\sqrt{a}-\sqrt{x}\bigr)^{2}$, $a\gt 0$, at an arbitrary point meets the axes at $A$ (on $Ox$) and $B$ (on $Oy$). Prove that $\overline{OA}+\overline{OB}$ is constant.

Step 1: The slope

The curve is defined for $x\ge0$. Differentiating, \[y' = 2\bigl(\sqrt{a}-\sqrt{x}\bigr)\cdot\left(-\frac{1}{2\sqrt{x}}\right) = -\frac{\sqrt{a}-\sqrt{x}}{\sqrt{x}}, \qquad k = -\frac{\sqrt{a}-\sqrt{x_{0}}}{\sqrt{x_{0}}} \tag{1}\] at a point $P = \bigl(x_{0},y_{0}\bigr)$ with $x_{0}\gt 0$ and $y_{0} = \bigl(\sqrt{a}-\sqrt{x_{0}}\bigr)^{2}$.

Step 2: The intercept on $Ox$

Setting $y = 0$ in $y-y_{0} = k\bigl(x-x_{0}\bigr)$ gives $x_{A} = x_{0}-\dfrac{y_{0}}{k}$, and by (1)

\[ \frac{y_{0}}{k} = \frac{\bigl(\sqrt{a}-\sqrt{x_{0}}\bigr)^{2}}{-\dfrac{\sqrt{a}-\sqrt{x_{0}}}{\sqrt{x_{0}}}} = -\sqrt{x_{0}}\bigl(\sqrt{a}-\sqrt{x_{0}}\bigr), \]

\[ x_{A} = x_{0}+\sqrt{ax_{0}}-x_{0} = \sqrt{ax_{0}} \;\Longrightarrow\; \overline{OA} = \sqrt{ax_{0}}. \tag{2} \]

Step 3: The intercept on $Oy$

Setting $x = 0$ gives $y_{B} = y_{0}-kx_{0}$, and since $kx_{0} = -\sqrt{ax_{0}}+x_{0}$, \[y_{B} = \bigl(a-2\sqrt{ax_{0}}+x_{0}\bigr)+\sqrt{ax_{0}}-x_{0} = a-\sqrt{ax_{0}} \;\Longrightarrow\; \overline{OB} = a-\sqrt{ax_{0}}. \tag{3}\]

Step 4: The sum

\[ \overline{OA}+\overline{OB} = \sqrt{ax_{0}}+a-\sqrt{ax_{0}} = a, \] independent of the point of contact. (The intercepts are taken as signed lengths: for $x_{0}\gt a$ the tangent meets $Oy$ below the origin and $\overline{OB}\lt 0$, the sum still being $a$.)

Step 5: Verification

\[a = 4,\ x_{0} = 1:\ \overline{OA} = 2,\ \overline{OB} = 2,\ \text{sum } 4 \;\;\checkmark\]

\[a = 4,\ x_{0} = 9:\ \overline{OA} = 6,\ \overline{OB} = -2,\ \text{sum } 4 \;\;\checkmark\]

\[x_{0} = a:\ \text{the curve touches } Ox,\ \overline{OA} = a,\ \overline{OB} = 0 \;\;\checkmark\]