If x - a t - 1t - y - a t - 1t where t be the parameter then dydx -

The correct option is C xy x = a (t 1t ), y = a (t + 1t ) y^2 x^2 = a^2 [ (t + 1t )^2 (t 1t )^2 ] y^2 x^2 = 4a^2 Differentiate ...

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Parametric Differentiation

If x = a t ...

Question

If $x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})$ where $t$ be the parameter then $\frac{d y}{d x} =$ ?

\frac{y}{x}

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\frac{- x}{y}

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\frac{x}{y}

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\frac{- y}{x}

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Solution

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Similar questions

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

\frac{d y}{d x} = \frac{x}{y}

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

\frac{d y}{d x} = \frac{x}{y}

A :

x = c t, y = \frac{c}{t}

t = 1, \frac{d y}{d x} =

B :

x = 3 cos θ - {cos}^{3} θ, y = 3 sin θ - {sin}^{3} θ

θ = \frac{π}{3}, \frac{d y}{d x} =

C :

x = a (t + \frac{1}{t}), y = a (t - \frac{1}{t})

t = 2, \frac{d y}{d x} =

D :

log (sec x)

tan x

x = \frac{π}{4}

x = \sqrt{a^{{sin}^{- 1} t}}, y = \sqrt{a^{{cos}^{- 1} t}}

\frac{d y}{d x} = - \frac{y}{x}

x^{2} + y^{2} = t - \frac{1}{t}

x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}}

x^{3} y \frac{d y}{d x}

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