If x-2 - y-2 - t - 1-t and x-4 - y-4 - t-2 - 1-t-2 -then x-3 y dy-dx equals- 101None of these

The correct option is C 1x^2 + y^2 = t 1/t and x^4 + y^4 = t_2 + 1/t^2 ∴ x^2 y^2 = 1 ⇒ x^2. 2y dy/dx + y^2 . 2x = 0 ⇒ x^3 y dy/dx = x^2 y^2 = 1 ...

If x^2 + y^2 ...

Question

If $x^{2} + y^{2} = t - \frac{1}{t}$ and $x^{4} + y^{4} = t_{2} + \frac{1}{t^{2}}$ ,then $x^{3} y \frac{d y}{d x}$ equals

– 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

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}

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}

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}

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} =

x^{2} + y^{2} = t + \frac{1}{t}, x^{4} + y^{4} = t^{2} + \frac{1}{t^{2}} \Rightarrow x^{3} y \frac{d y}{d x} =