First slide

Extreme values and Periodicity of Trigonometric functions

Question

$The minimum distance of the curve \frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}} = 1 from origin is (a, b > 0)$

Moderate

A

$|a - b|$
B

$a + b$
C

$a b$
D

0

Solution

$Given curve is \frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}} = 1 (h e r e x \leq - a o r x \geq a & y ⩽ - b o r y ⩾ b) compare with \cos^{2} θ + \sin^{2} θ = 1$
$\begin{array}{l} \Rightarrow \frac{a}{x} = \cos θ, \frac{b}{y} = \sin θ \\ \Rightarrow x = a \sec θ and y = b cosec θ \\ Minimum distance of the curve from the origin = \sqrt{x^{2} + y^{2}} \end{array}$
$\begin{array}{l} = \sqrt{a^{2} \sec^{2} θ + b^{2} {cosec}^{2} θ} \\ = \sqrt{a^{2} + b^{2} + a^{2} \tan^{2} θ + b^{2} \cot^{2} θ} \\ = \sqrt{a^{2} + b^{2} + (a \tan θ - b \cot θ)^{2} + 2 a b} \\ = \sqrt{(a + b)^{2} + (a \tan θ - b \cot θ)^{2}} \\ \geq a + b \\ Therefore, the correct answer is (2). \end{array}$

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