The gradient of \(L\) must satisfy that \begin {align*} \frac {\Diff y}{\Diff x} & = \frac {\Diff y / \Diff t}{\Diff x / \Diff t} \\ & = \frac {b}{a} \cdot \frac {\Diff \left (2t/(1+t^2)\right ) / \Diff t}{\Diff \left ((1-t^2)/(1+t^2)\right ) / \Diff t} \\ & = \frac {b}{a} \cdot \frac {2 \cdot (1+t^2) - 2t \cdot 2t}{-2t \cdot (1+t^2) - (1-t^2) \cdot 2t} \\ & = \frac {b}{a} \cdot \frac {2 + 2t^2 - 4t^2}{-2t - 2t^3 - 2t + 2t^3} \\ & = \frac {b}{a} \cdot \frac {1 - t^2}{-2t}. \end {align*}
Therefore, we have a general point \((X, Y) \in L\) satisfy that \begin {align*} Y - \frac {2bt}{1+t^2} & = \frac {b}{a} \cdot \frac {1 - t^2}{-2t} \cdot \left (X - \frac {a(1-t^2)}{1+t^2}\right ) \\ (1+t^2)Y - 2bt & = \frac {b}{a} \cdot \frac {1-t^2}{-2t} \cdot \left ((1+t^2)X - a(1-t^2)\right ) \\ (-2at)(1+t^2)Y - (-2at)(2bt) & = b \cdot (1-t^2) \cdot \left ((1+t^2)X - a(1-t^2)\right ) \\ (-2at)(1+t^2)Y & = b(1-t^2)(1+t^2)X - ab(1-t^2)^2 - 4abt^2 \\ (-2at)(1+t^2)Y & = b(1-t^2)(1+t^2)X - ab(1+t^2)^2 \\ -2atY & = b(1-t^2)X - ab(1+t^2) \\ ab(1+t^2) -2atY - b(1-t^2)X & = 0 \\ (a + X)bt^2 - 2aYt + b(a-X) & = 0 \end {align*}
as desired.
Now if we fix \(X, Y\) and solve for \(t\), there are two solutions to this quadratic equation exactly when \begin {align*} (2aY)^2 - 4(a+X)b \cdot b(a-X) & >0 \\ (aY)^2 - (a+X)(a-X)b^2 & >0 \\ a^2Y^2 & > (a^2-X^2)b^2, \end {align*}
which corresponds to two distinct points on the ellipse.
Since \(a^2 Y^2 > (a^2 - X^2) b^2\), we have \(\frac {Y^2}{b^2} > 1 - \frac {X^2}{a^2}\) by dividing through \(a^2 b^2\) on both sides, i.e. \[ \frac {X^2}{a^2} + \frac {Y^2}{b^2} > 1, \] which means when the point \((X, Y)\) lies outside the ellipse.
This also holds when \(X^2 = a^2\), i.e. when the point \((X, Y)\) lies on the pair of lines \(X = \pm A\). Here, the condition is simply \(a^2 Y^2 > 0\), which gives \(Y \neq 0\). One of the tangents will be the vertical line \(X = \pm A\) (whichever one the point lies on), and the other one as a non-vertical (as shown when \(X = a\), the tangents being \(L_1\) and \(L_2\)).
By Vieta’s Theorem, we have \[ pq = \frac {b(a-X)}{b(a+X)} \implies (a+X) pq = a-X, \] as desired, and \[ p + q = -\frac {-2aY}{(a+X)b} = \frac {2aY}{(a+X)b}. \]
Let \(X = 0\) for the equation in \(L\), \begin {align*} abt^2 - 2aYt + ba & = 0 \\ bt^2 - 2Yt + b & = 0 \\ Y & = \frac {b(1+t^2)}{2t}. \end {align*}
Therefore, \begin {align*} y_1 + y_2 & = \frac {b(1+p^2)}{2p} + \frac {b(1+q^2)}{2q} \\ & = \frac {b\left [(1+p^2)q + (1+q^2)p\right ]}{2pq} \\ & = 2b, \end {align*}
therefore we have \[ 4pq = (1+p^2)q + (1+q^2)p = (p+q)(1 + pq). \]
Therefore, \begin {align*} 4 \cdot \frac {a-X}{a+X} & = \frac {2aY}{(a+X)b} \cdot \frac {2a}{a+X} \\ a-X & = \frac {a^2 Y}{b(a+X)} \\ (a-X)(a+X)b & = a^2Y \\ (a^2 - X^2) b & = a^2 Y \\ 1 - \frac {X^2}{a^2} & = \frac {Y}{b} \\ \frac {X^2}{a^2} + \frac {Y}{b} & = 1, \end {align*}
as desired.