Question

If \[ \frac{y}{x}\cos^4\alpha+\frac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha, \] then \[ \frac{dy}{dx}= \]

• A. \(\sin^3\alpha\cos\alpha\)
• B. \(\sin^2\alpha\cos^2\alpha\)
• C. \(\dfrac{\sin^2\alpha}{\cos^2\alpha}\)
• D. \(\sin\alpha\cos^3\alpha\)

Hint:

When an equation contains \(\frac{y}{x}\) and \(\frac{x}{y}\), put \(\frac{y}{x}=t\) to simplify it quickly.

Answer 1

The Correct Option is C

Step 1: Put \(\frac{y}{x}=t\).
Given: \[ \frac{y}{x}\cos^4\alpha+\frac{x}{y}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha \] Let \[ t=\frac{y}{x} \] Then, \[ \frac{x}{y}=\frac{1}{t} \] So, \[ t\cos^4\alpha+\frac{1}{t}\sin^4\alpha = 2\sin^2\alpha\cos^2\alpha \]

Step 2: Multiply by \(t\).
\[ t^2\cos^4\alpha+\sin^4\alpha = 2t\sin^2\alpha\cos^2\alpha \] Bring all terms to one side: \[ t^2\cos^4\alpha-2t\sin^2\alpha\cos^2\alpha+\sin^4\alpha=0 \] This is a perfect square: \[ (t\cos^2\alpha-\sin^2\alpha)^2=0 \] Therefore, \[ t\cos^2\alpha=\sin^2\alpha \] \[ t=\frac{\sin^2\alpha}{\cos^2\alpha} \]

Step 3: Find relation between \(x\) and \(y\).
Since \[ t=\frac{y}{x}, \] we get \[ \frac{y}{x}=\frac{\sin^2\alpha}{\cos^2\alpha} \] Thus, \[ y=x\frac{\sin^2\alpha}{\cos^2\alpha} \]

Step 4: Differentiate with respect to \(x\).
Since \(\alpha\) is constant, \[ \frac{dy}{dx} = \frac{\sin^2\alpha}{\cos^2\alpha} \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac{\sin^2\alpha}{\cos^2\alpha}} \]