(C) Given equations are $x = a (t - 1/t)$ and $y = a (t + 1/t)$.Squaring and subtracting the equations:$y^2 - x^2 = a^2 [(t + 1/t)^2 - (t - 1/t)^2]$Us

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(C) Given equations are $x = a (t - 1/t)$ and $y = a (t + 1/t)$.Squaring and subtracting the equations:$y^2 - x^2 = a^2 [(t + 1/t)^2 - (t - 1/t)^2]$Using the identity $(a+b)^2 - (a-b)^2 = 4ab$,we get:$y^2 - x^2 = a^2 [4 \cdot t \cdot (1/t)] = 4a^2$.Differentiating both sides with respect to $x$:$d/dx (y^2 - x^2) = d/dx (4a^2)$$2y (dy/dx) - 2x = 0$$2y (dy/dx) = 2x$Therefore,$dy/dx = x/y$.

Class 12 Mathematics Continuity and Differentiability Derivatives of Functions in Parametric Forms

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If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$

MHT CET 2017 All MHT CET

A
$y/x$
B
$-x/y$
C
$x/y$
D
$-y/x$

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Continuity and Differentiability

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Derivatives of Functions in Parametric Forms

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If $x = a (t - 1/t)$ and $y = a (t + 1/t)$,where $t$ is the parameter,then $dy/dx = ?$

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