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

• A. a + b
• B. a - b
• C. $a^2+b^2$
• D. $a^2-b^2$

Answer 1

The correct option is A a + b

Equation of tangent at $\left(\theta \right)$ is
$\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1ItmeetsaxesatM=(\frac{a}{cos\theta },0)andN=(0,\frac{b}{sin\theta })\therefore MN=\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }=\sqrt{a^2+b^2+a^{2}co{t}^2\theta +b^{2}ta{n}^2\theta }\therefore Minimumvalueof{a}^{2}co{t}^2\theta +b^{2}ta{n}^2\theta is2ab(\because A.M\ge G.M)\therefore MinimumvalueofMN=a+b$