Q.3 (B) (ii) If x=4t/1+t² , y=3(1-t²/1+t²) , then show that dy/dx=-9x/4y .

Answer: Proved below. dx/dt=4·(1+t²)(1)-t(2t)/(1+t²)²=4(1-t²)/(1+t²)² dy/dt=3·(1+t²)(-2t)-(1-t²)(2t)/(1+t²)²=3·-4t/(1+t²)²=-12t/(1+t²)² dy/dx=dy/dt/dx/dt=-12t/4(1-t²)=-3t/1-t² ... (1) Now -9x/4y=-9·4t/1+t²/4·3·1-t²/1+t²=-36t/12(1-t²)=-3t/1-t² ... (2) From (1) and (2), dy/dx=-9x/4y .

If \(x=\frac{4t}{1+t^2}\), \(y=3\left(\frac{1-t^2}{1+t^2}\right)\), then show that \(\frac{dy}{dx}=\frac{-9x}{4y}\).

Q.3 (B) (ii) If \(x=\frac{4t}{1+t^2}\), \(y=3\left(\frac{1-t^2}{1+t^2}\right)\), then show that \(\frac{dy}{dx}=\frac{-9x}{4y}\).

Doubtly Asked 28 Oct 2026 Board exam · March 2025 hsc 12th Save WhatsApp Report

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Answer: Proved below.

\(\frac{dx}{dt}=4\cdot\frac{(1+t^2)(1)-t(2t)}{(1+t^2)^2}=\frac{4(1-t^2)}{(1+t^2)^2}\)

\(\frac{dy}{dt}=3\cdot\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}=3\cdot\frac{-4t}{(1+t^2)^2}=\frac{-12t}{(1+t^2)^2}\)

\(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-12t}{4(1-t^2)}=\frac{-3t}{1-t^2}\) ... (1)

Now \(\frac{-9x}{4y}=\frac{-9\cdot\frac{4t}{1+t^2}}{4\cdot3\cdot\frac{1-t^2}{1+t^2}}=\frac{-36t}{12(1-t^2)}=\frac{-3t}{1-t^2}\) ... (2)

From (1) and (2), \(\frac{dy}{dx}=\frac{-9x}{4y}\).

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If \(x=\frac{4t}{1+t^2}\), \(y=3\left(\frac{1-t^2}{1+t^2}\right)\), then show that \(\frac{dy}{dx}=\frac{-9x}{4y}\).

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