Find the point(s) on the ellipse: x²/a² + (y - k)²/b² = 1 where the tangent lines that passes through the point: (m, n)?

Since linear transformations leave incidence properties (and hence tangents) invariant, translate "k" units down, then compress vertically by "1/b" and horizontally by "1/a". That takes the ellipse to a unit circle about the origin, x²+y²=1 and the point (m,n) to (m/a,(n-k)/b)=(M,N). {Note that M²+N²≥1 is the same as m²/a²+(n-k)²/b²≥1.}

Now look at the similar right triangles formed by the origin O, P(x,y), Q(M,N), and the orthogonal projection of (x,y) on the line from the origin to (M,N), calling that R. A little playing around with similar triangles gives|OM|/|OX|=|OX|/|OR|, and |RX|/|OX| = |MX|/|OM| = √(|OM|²-|OX|²)/|OM|. Since |OX|=1, |OR|=1/|OM| and |RX|=&radic(|OM|²-1)/|OM|. Furthermore, |RX| is (by construction) perpendicular to |OM|, hence it is parallel to (N, -M). Therefore:

        (RX) = ±√(|OM|²-1)/|OM|² (N, -M) = ±√(M²+N²-1)/(M²+N²) (N, -M)

and

        (OR) = (M, N) ÷ |OM|² = (M, N) / (M²+N²)

so

        (x,y) = (OR) + (RX) = (M±N√(M²+N²-1), N∓M√(M²+N²-1)) / (M²+N²)

{Note the requirement that M²+N²≥1. The point (m,n) must be outside the ellipse.}

Decompressing and then translating up produces the solution to the original problem:

        (x,y) = (aM ± aN√(M²+N²-1),  bN ∓ bM√(M²+N²-1) + k(M²+N²)) ÷ (M²+N²)

        = (m ± (a/b)(n-k)√Δ,  n ∓ (b/a)m√Δ + kΔ) / (Δ+1),   where   Δ = m²/a² + (n-k)²/b² - 1 ≥ 0

Check my algebra! Even this simplied approach gave me trouble...

UPDATE: Thanks for showing me your solution! I think I like my solution. In addition to being algebraically easier, this one also works for an ellipse with a horizontal translation (h≠0). In fact, generalizing to the real projective plane, this idea works for any conic. In the projective plane any conic can be transformed to a unit circle with a linear transformation, where the problem is solved with similar triangles and one application of the Pythagorean theorem. The inverse linear transforation gets the desired solutions to the original problem.

Specifically, for h≠0, we have M=(m-h)/a and N=(n-k)/b, so Δ = (m-h)²/a² + (n-k)²/b² - 1

        x = [ m ± (a/b)(n-k)√Δ + hΔ ] ÷ (Δ+1)

        y = [ n ∓ (b/a)(m-h)√Δ + kΔ ] ÷ (Δ+1)

Which reduces to your answer in the case h=0.

Here's my work:

https://plus.google.com/photos/1151511903526799645...