If x = 1 + t 2 1 − t 2 and y = 1 + t 2 2 t , then d x d y is equal to :

Let x = 1 + t 2 1 − t 2 and y = 1 + t 2 2 t Put t = tan θ , we get x = 1 + t a n 2 θ 1 − t a n 2 θ and y = 1 + t a n 2 θ 2 t a n θ ⇒ x = cos 2 θ and y = sin 2 θ ∴ d θ dx = − 2 sin 2 θ and d θ dy = 2 cos 2 θ Now, dx dy = d θ dy × dx d θ = − s i n 2 θ c o s 2 θ = − y x

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MathematicsDifferentiationParametric/Determinant/Logarithmic InverseMedium2 minPYQ_2021

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If $x = \frac{1 - t ^{2}}{1 + t ^{2}}$ and $y = \frac{2 t}{1 + t ^{2}}$ , then $\frac{d y}{d x}$ is equal to :

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Let $x = \frac{1 - t ^{2}}{1 + t ^{2}}$ and $y = \frac{2 t}{1 + t ^{2}}$ Put $t = tan θ$ , we get $x = \frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ}$ and $y = \frac{2 t a n θ}{1 + t a n ^{2} θ}$ $\Rightarrow x = cos 2 θ$ and $y = sin 2 θ$ $∴ \frac{dx}{d θ} = - 2 sin 2 θ$ and $\frac{dy}{d θ} = 2 cos 2 θ$ Now, $\frac{dy}{dx} = \frac{dy}{d θ} \times \frac{d θ}{dx} = - \frac{c o s 2 θ}{s i n 2 θ} = - \frac{x}{y}$

Stream:BITSATSubject:MathematicsTopic:DifferentiationSubtopic:Parametric/Determinant/Logarithmic Inverse

⏱ 2mℹ️ Source: PYQ_2021

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