If x = sec ⁡ t + tan ⁡ t and y = sec ⁡ t - tan ⁡ t , where t is a parameter, then the value of d y d x when x = 1 3 is

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Question: Answered & Verified by Expert

If

x = \sec t + \tan t

and

y = \sec t - \tan t

, where

t

is a parameter, then the value of

\frac{d y}{d x}

when

x = \frac{1}{\sqrt{3}}

is

MathematicsDifferentiationJEE Main

Options:

A $0$
B $- 3$
C $\sqrt{3}$
D $\frac{1}{\sqrt{3}}$

Solution:

2656 Upvotes Verified Answer

The correct answer is:

- 3

∵ x y = 1 \Rightarrow y = \frac{1}{x}

\frac{d y}{d x} = \frac{- 1}{x^{2}}

at

x = \frac{1}{\sqrt{3}}, \frac{d y}{d x} = - 3

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