If the minimum area of the triangle formed by a tangent to the ellipse x^2/b^2 + y^2/4a^2 = 1and the co-ordinate axis is kab,

Question

If the minimum area of the triangle formed by a tangent to the ellipse \(\frac{x^2}{b^2}\) + \(\frac{y^2}{4a^2}\) = 1 and the co-ordinate axis is kab, then k is equal to _______.

x²/b² + y²/4a² = 1

Answer 1

Tangent

\(\frac{xcos\theta}{b}\) + \(\frac{ysin\theta}{2a}\) = 1

So, 

area(ΔOAB) = \(\frac{1}{2}\) x \(\frac{b}{cos\theta}\) x \(\frac{2a}{sin\theta}\)

= \(\frac{2ab}{sin2\theta}\) ≥ 2ab

⇒ k = 2