A tangent to the ellipse x2a2+y2b2=1 cuts the axes in M and N. Then the least le

Question

A tangent to the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

cuts the axes in M and N. Then the least length of MN is

Options:

A . a + b

B . a - b

C . a2+b2

D . a2−b2

Answer: Option A
:
A
Equation of tangent at

(θ)

\frac{x}{a} c o s θ + \frac{y}{b} s i n θ = 1 I t m e e t s a x e s a t M = (\frac{a}{c o s θ}, 0) a n d N = (0, \frac{b}{s i n θ}) ∴ M N = \sqrt{a^{2} s e c^{2} θ + b^{2} c o s e c^{2} θ} = \sqrt{a^{2} + b^{2} + a^{2} c o t^{2} θ + b^{2} t a n^{2} θ} ∴ M i n i m u m v a l u e o f a^{2} c o t^{2} θ + b^{2} t a n^{2} θ i s 2 a b (∵ A . M \geq G . M) ∴ M i n i m u m v a l u e o f M N = a + b

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