Question

Statement -1 : The minimum sum of intercepts that a tangent to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ make between the coordinate axes is $a+b.$
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

Answer: (C)

Statement-1 is True, 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 $b\text{cosec}\theta$

$\therefore$ Length of intercept $=\sqrt{a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta }$
$a^2{\sec }^2\theta +b^2{\text{cosec}}^2\theta =a^2+b^2+a^2{\tan }^2\theta +b^2{\cot }^2\theta \ge (a+b{)}^2$

So the minimum value of length is $(a+b)$

Statement 2:

It says that $a$ and $b$ are negative real numbers.
So let us substitute $a=-x$ and $b=-y$ where $x,y$ $\in$ $R^{+}$
Thus, equation becomes $\frac{-(x+y)}{2}\ge -\sqrt{xy}$
Thus, $\frac{x+y}{2}\le \sqrt{xy}$
If squared on both sides, we get $(\sqrt{x}+\sqrt{y}{)}^2\le 0.$

That's not true and hence option C.