Question

Assertion :STATEMENT-1 : The minimum value of the intercepts cut on tangent by a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ between the coordinate axes is a + b Reason: STATEMENT-2 : For each pair of two negative real numbers a and b, inequality $\frac{a+b}{2}\ge -\sqrt{ab}$ holds.

• A. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
• B. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
• C. Statement-1 is True, Statement-2 is False
• D. Statement-1 is False, Statement-2 is True

Answer 1

The correct option is C Statement-1 is True, Statement-2 is False

STATEMENT- 2 Let a = -a', b = -b'

$\therefore \frac{a+b}{2}=-\left[\frac{a^{\prime }+b^{\prime }}{2}\right]\le -\sqrt{a^{\prime }{b}^{\prime }}=-\sqrt{ab}$

$\therefore$ STATEMENT - 2 is false.

STATEMENT - 1 Equation of tangent is $x\frac{cos\theta }{a}+y\frac{sin\theta }{b}=1$

$\therefore$ The intercepts are a $sec\theta$ and $bcosec\theta$

$\therefore$ Length of intercept =$\sqrt{a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta }$

$a^{2}se{c}^2\theta +b^{2}cose{c}^2\theta =a^2+b^2+a^{2}ta{n}^2\theta +b^{2}co{t}^2\theta \ge (a+b{)}^2$