Question

Find the locus of the point of intersection of the tangents to the ellipse:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
(a > b), if the difference of the eccentric angles of their points of contact is 2*α.

Ans:
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=sec^2 \alpha
[/tex]

Answer 1

A point (x,y) on the ellipse is:
    x = a cos t, y = b sin t,
    t = eccentric angle at (x,y) on the ellipse.  t = tan⁻¹ (ya /bx)

Tangent at (a cos t, b sin t):
        y = m x +-  sqrt(a^2 m^2 + b^2)
     Or,   x cos t /a + y sin t /b = 1,   slope m = (-b/a) cot t        ---(1)

Tangent at {a cos(t+2α), b sin(t+2α)}:
        x cos(t+2α) /a + y sin(t+2α) /b = 1      --- (2)

Point of intersection of the two tangents:
    x [cos(t+2α) - cos t] /a + y [sin(t+2α) - sin t] /b = 0
    -(2 x/a) sin(t+α) sin α + (2 y/b) sin α cos(t+α) = 0
      y = x  (b/a) Tan (t +α) 

Substitute for y in (1) to get :
       x = a/[cos t + Tan(t +α) sin t] = a cos(t+α)/cos α   -- (3)
       y = b Sin (t +α) / cos α  -- (4)

Eliminating t from (3) and (4),
     x²/a² +y²/b² =  sec² α