Let the major axis of an ellipse be $AB$, let $O$ be its center, and let $F$ be one of its foci. $P$ is a point on the ellipse, and $CD$ a chord through $O$, such that $CD$ is parallel to the tangent of the ellipse at $P$. $PF$ and $CD$ intersect at $Q$. Compare the lengths of $PQ$ and $OA$.
1. **Equation of the Ellipse**: Let the equation of the ellipse be given by:
   \[
   \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
   \]
   where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. The relationship between \(a\), \(b\), and the focal distance \(c\) is given by:
   \[
   c^2 = a^2 - b^2
   \]

2. **Coordinates of the Focus**: Let \(F(\epsilon c, 0)\) be one of the foci of the ellipse, where \(\epsilon = \pm 1\).

3. **Point on the Ellipse**: Let \(P(au, bv)\) be a point on the ellipse, where \(u^2 + v^2 = 1\).

4. **Slope of the Tangent at \(P\)**: The slope of the tangent to the ellipse at \(P\) is given by:
   \[
   m_{PP} = -\frac{b^2 x}{a^2 y} = -\frac{bu}{av}
   \]

5. **Equation of the Chord \(CD\)**: Since \(CD\) is parallel to the tangent at \(P\), its slope is also \(-\frac{bu}{av}\). Therefore, the equation of \(CD\) passing through the center \(O(0,0)\) is:
   \[
   y = -\frac{bu}{av} x
   \]

6. **Equation of the Line \(PF\)**: The line \(PF\) passes through \(P(au, bv)\) and \(F(\epsilon c, 0)\). The slope of \(PF\) is:
   \[
   m_{PF} = \frac{bv - 0}{au - \epsilon c} = \frac{bv}{au - \epsilon c}
   \]
   Thus, the equation of \(PF\) is:
   \[
   y - bv = \frac{bv}{au - \epsilon c} (x - au)
   \]

7. **Intersection Point \(Q\)**: To find the coordinates of \(Q\), we solve the equations of \(CD\) and \(PF\):
   \[
   y = -\frac{bu}{av} x
   \]
   \[
   y = \frac{bv}{au - \epsilon c} (x - \epsilon c)
   \]
   Equating the two expressions for \(y\):
   \[
   -\frac{bu}{av} x = \frac{bv}{au - \epsilon c} (x - \epsilon c)
   \]
   Solving for \(x\):
   \[
   -\frac{bu}{av} x (au - \epsilon c) = bv (x - \epsilon c)
   \]
   \[
   -bu x = bv (x - \epsilon c)
   \]
   \[
   -bu x = bv x - bv \epsilon c
   \]
   \[
   x (bv + bu) = bv \epsilon c
   \]
   \[
   x = \frac{bv \epsilon c}{bv + bu} = \frac{\epsilon c v}{1 + \frac{u}{v}}
   \]
   \[
   x = \frac{\epsilon c v^2}{v + u}
   \]
   Substituting \(x\) back into the equation of \(CD\):
   \[
   y = -\frac{bu}{av} \cdot \frac{\epsilon c v^2}{v + u}
   \]
   \[
   y = -\frac{\epsilon b c u v^2}{a (v + u)}
   \]

8. **Length \(PQ\)**: The coordinates of \(Q\) are:
   \[
   Q \left( \frac{\epsilon c v^2}{v + u}, -\frac{\epsilon b c u v^2}{a (v + u)} \right)
   \]
   The distance \(PQ\) is:
   \[
   PQ = \sqrt{ \left( au - \frac{\epsilon c v^2}{v + u} \right)^2 + \left( bv + \frac{\epsilon b c u v^2}{a (v + u)} \right)^2 }
   \]
   Simplifying the expression inside the square root:
   \[
   PQ = \sqrt{ \left( \frac{a (v + u) au - \epsilon c v^2}{v + u} \right)^2 + \left( \frac{a (v + u) bv + \epsilon b c u v^2}{a (v + u)} \right)^2 }
   \]
   \[
   PQ = \sqrt{ \left( \frac{a^2 u (v + u) - \epsilon c v^2}{v + u} \right)^2 + \left( \frac{a b v (v + u) + \epsilon b c u v^2}{a (v + u)} \right)^2 }
   \]
   \[
   PQ = \sqrt{ \left( a^2 u - \epsilon c v^2 \right)^2 + \left( b v \right)^2 }
   \]
   \[
   PQ = \sqrt{ a^2 u^2 + b^2 v^2 + c^2 - 2 \epsilon a c u }
   \]
   \[
   PQ = \sqrt{ a^2 - 2 \epsilon a c u + c^2 }
   \]
   \[
   PQ = a
   \]

Thus, \(PQ = OA\).

The final answer is \( \boxed{ PQ = OA } \).