The tangent to y = (a²)/2x at an arbitrary point meets

The tangent to $y = \dfrac{a^{2}}{2x}$ at an arbitrary point meets the axes at $A$ and $B$. 1° Show that the area of $\triangle OAB$ is constant and equal to $a^{2}$. 2° Find the distance from $O$ to the tangent.

Step 1: The tangent

At $P = \left(x_{0},\dfrac{a^{2}}{2x_{0}}\right)$, $x_{0}\neq0$, the slope is $y' = -\dfrac{a^{2}}{2x^{2}}$, so

\[ y = -\frac{a^{2}}{2x_{0}^{2}}x+\frac{a^{2}}{2x_{0}}+\frac{a^{2}}{2x_{0}} = -\frac{a^{2}}{2x_{0}^{2}}x+\frac{a^{2}}{x_{0}}. \tag{1} \]

Step 2: The intercepts

\[ y = 0 \;\Longrightarrow\; x = 2x_{0}, \qquad A = \bigl(2x_{0},0\bigr); \qquad x = 0 \;\Longrightarrow\; B = \left(0,\frac{a^{2}}{x_{0}}\right). \tag{2} \]

The point of contact is the midpoint of $AB$ — a property of every hyperbola $xy = \text{const}$.

Step 3: 1° — the area

The triangle has its right angle at $O$, with legs $|OA|$ and $|OB|$: \[S = \frac12\bigl|2x_{0}\bigr|\cdot\left|\frac{a^{2}}{x_{0}}\right| = \frac12\cdot2\bigl|x_{0}\bigr|\cdot\frac{a^{2}}{\bigl|x_{0}\bigr|} = a^{2},\] the same for every point of contact.

Step 4: 2° — the distance from the origin

Writing (1) as $\dfrac{a^{2}}{2x_{0}^{2}}x+y-\dfrac{a^{2}}{x_{0}} = 0$ and using $d = \dfrac{|\gamma|}{\sqrt{\alpha^{2}+\beta^{2}}}$: \[d = \frac{\dfrac{a^{2}}{\bigl|x_{0}\bigr|}}{\sqrt{\dfrac{a^{4}}{4x_{0}^{4}}+1}} = \frac{\dfrac{a^{2}}{\bigl|x_{0}\bigr|}}{\dfrac{\sqrt{a^{4}+4x_{0}^{4}}}{2x_{0}^{2}}} = \frac{2a^{2}\bigl|x_{0}\bigr|}{\sqrt{a^{4}+4x_{0}^{4}}}. \tag{3}\]

Step 5: Verification

\[\text{Intercept form } \frac{x}{2x_{0}}+\frac{y}{a^{2}/x_{0}} = 1 \;\Longrightarrow\; d = \frac{1}{\sqrt{\dfrac{1}{4x_{0}^{2}}+\dfrac{x_{0}^{2}}{a^{4}}}} = \frac{2a^{2}|x_{0}|}{\sqrt{a^{4}+4x_{0}^{4}}} \;\;\checkmark\]

\[a = 2,\ x_{0} = 1:\ A = \bigl(2,0\bigr),\ B = \bigl(0,4\bigr),\ S = 4 = a^{2},\ d = 1.7888544 \;\;\checkmark\]

\[\text{Maximum of } d:\ \text{at } x_{0}^{2} = \frac{a^{2}}{2},\ d = a\ \text{(the vertex of the hyperbola)} \;\;\checkmark\]