Q.40. Ellipse

Question 40.

Find the point of intersection of the tangents drawn to ellipse x^2/a^2 + y^2/b^2 = 1 at two points whose eccentric angles are A-B and A+B.

Answer 40.

The equation of tangent at an eccentric angle θ to the ellipse x^2/a^2 + y^2/b^2 = 1 is

xcosθ/a + ysinθ/b = 1, i.e.,

(bcosθ)x + (asinθ)y - ab = 0.

 The equations of tangents at the eccentric angles A-B and A+B are
[bcos(A-B)]x + [asin(A-B)]y - ab = 0 and
[bcos(A+B)]x + [asin(A+B)]y - ab = 0

Solving these equations simultaneously using Cramer's formula,
x = a^2b[sin(A+B) - sin(A-B)] / ab[sin(A+B)cos(A-B) - cos(A+B)sin(A-B)] = acosA/cosB and
y = ab^2[cos(A-B) - cos(A+B)] / ab[sin(A+B)cos(A-B) - cos(A+B)sin(A-B)] = bsinA/cosB

=> the point of intersection of the tangents is
(acosA/cosB, bsinA/cosB).

 LINK to YA!