Question

If $x^{2}+y^{2}=t+\frac{1}{t}$ and $x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{y}{x}$
• B. $\frac{-y}{x}$
• C. $\frac{x}{y}$
• D. $\frac{-x}{y}$

Hint:

Logic Tip: Alternatively, from $x^2y^2 = 1$, we get $xy = 1$ (for 1st quadrant). Differentiating gives $x\frac{dy}{dx} + y = 0 \implies \frac{dy}{dx} = -\frac{y}{x}$. Taking the root before differentiating simplifies the product rule application!

Answer 1

The Correct Option is B

Concept:
Eliminate the parameter $t$ by squaring the first equation to match the terms of the second equation. Once $t$ is eliminated, implicitly differentiate the resulting relation between $x$ and $y$ to find $\frac{dy}{dx}$.
Step 1: Square the first equation to eliminate parameter t.
Given: $$x^2 + y^2 = t + \frac{1}{t}$$ Square both sides: $$(x^2 + y^2)^2 = \left(t + \frac{1}{t}\right)^2$$ $$x^4 + y^4 + 2x^2y^2 = t^2 + \frac{1}{t^2} + 2\left(t \cdot \frac{1}{t}\right)$$ $$x^4 + y^4 + 2x^2y^2 = t^2 + \frac{1}{t^2} + 2$$
Step 2: Substitute the second equation into the expanded form.
We are given $x^4 + y^4 = t^2 + \frac{1}{t^2}$. Substitute this into our expanded equation: $$(t^2 + \frac{1}{t^2}) + 2x^2y^2 = t^2 + \frac{1}{t^2} + 2$$ Subtract $(t^2 + \frac{1}{t^2})$ from both sides: $$2x^2y^2 = 2$$ $$x^2y^2 = 1$$
Step 3: Implicitly differentiate to find dy/dx.
Differentiate the equation $x^2y^2 = 1$ with respect to $x$ using the product rule: $$\frac{d}{dx}(x^2 \cdot y^2) = \frac{d}{dx}(1)$$ $$x^2 \cdot \frac{d}{dx}(y^2) + y^2 \cdot \frac{d}{dx}(x^2) = 0$$ $$x^2(2y \cdot \frac{dy}{dx}) + y^2(2x) = 0$$ Divide by 2: $$x^2y \frac{dy}{dx} + xy^2 = 0$$ Isolate $\frac{dy}{dx}$: $$x^2y \frac{dy}{dx} = -xy^2$$ $$\frac{dy}{dx} = \frac{-xy^2}{x^2y} = -\frac{y}{x}$$