If the normals to x^2/a^2 + y^2/b^2 = 1 at the ends of the chords l1x + m1y =1 and l2x + m2y =1 are concurrent, then:

Question

If the normals to \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at the ends of the chords l₁x + m₁y =1 and l₂x + m₂y =1 are concurrent, then:

(a) a²l₁l₂ + b²m₁m₂ = 1

(b) a²l₁l₂ + b₂m₁m₂ = -1

(c) a²l₁l₂ - b²m₁m₂ = 1

(d) None of these

Answer 1

(b) a²l₁l₂ + b₂m₁m₂ = -1