JEE Advance - Mathematics (1997 - No. 19)
A tangent to the ellipse x2 + 4y2 = 4 meets the ellipse x2 + 2y2 = 6 at P and Q. Prove that the tangents at P and Q of the ellipse x2 + 2y2 = 6 are at right angles.
The equation of a tangent to the ellipse x2 + 4y2 = 4 is y = mx + sqrt(1 + 4m2)
The equation of a tangent to the ellipse x2 + 4y2 = 4 is y = mx + sqrt(4 + m2)
The equation of a tangent to the ellipse x2 + 4y2 = 4 is y = mx + sqrt(1 + m2)
Substituting the tangent equation into x2 + 2y2 = 6 gives a quadratic equation in x. The product of roots condition for perpendicularity is used.
Substituting the tangent equation into x2 + 2y2 = 6 gives a quadratic equation in x. The sum of roots condition for perpendicularity is used.