Approximated formula of acid dissociation constant
The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.

We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.

We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.

Here is some mathematics that you requested:


  $K_{\mathrm{a}}$ can be calculated using equilibrium concentrations:
  $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$


Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)



Here is an extension to fully show how the two equate. Taken from the same textbook:


  $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$
  Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$