In a right triangle with sides $ a$ and $ b$, and hypotenuse $ c$, the altitude drawn on the hypotenuse is $ x$. Then:

$ \textbf{(A)}\ ab \equal{} x^2 \qquad\textbf{(B)}\ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{x} \qquad\textbf{(C)}\ a^2 \plus{} b^2 \equal{} 2x^2$
$ \textbf{(D)}\ \frac {1}{x^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \qquad\textbf{(E)}\ \frac {1}{x} \equal{} \frac {b}{a}$
1. Given a right triangle with sides \(a\) and \(b\), and hypotenuse \(c\), we know from the Pythagorean theorem that:
   \[
   c^2 = a^2 + b^2
   \]

2. Let \(x\) be the altitude drawn to the hypotenuse \(c\). By a well-known geometric property of right triangles, the product of the segments of the hypotenuse created by the altitude is equal to the product of the legs:
   \[
   a \cdot b = c \cdot x
   \]

3. Solving for \(x\), we get:
   \[
   x = \frac{a \cdot b}{c}
   \]

4. Squaring both sides, we obtain:
   \[
   x^2 = \left(\frac{a \cdot b}{c}\right)^2 = \frac{a^2 b^2}{c^2}
   \]

5. Substituting \(c^2 = a^2 + b^2\) into the equation, we get:
   \[
   x^2 = \frac{a^2 b^2}{a^2 + b^2}
   \]

6. To find \(\frac{1}{x^2}\), we take the reciprocal of \(x^2\):
   \[
   \frac{1}{x^2} = \frac{a^2 + b^2}{a^2 b^2}
   \]

7. This can be rewritten as:
   \[
   \frac{1}{x^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}
   \]

8. Therefore, we have:
   \[
   \frac{1}{x^2} = \frac{1}{a^2} + \frac{1}{b^2}
   \]

9. This matches option \(\text{D}\).

The final answer is \(\boxed{\frac{1}{x^2} = \frac{1}{a^2} + \frac{1}{b^2}}\)