Question

If \( (a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y) = a^2 - b^2 \), where \( a>b>0 \), then at \( \left( \dfrac{\pi}{4}, \dfrac{\pi}{4} \right) \), \( \dfrac{dy}{dx} = \)

• A. \( \dfrac{a + b}{a - b} \)
• B. \( \dfrac{a - b}{a + b} \)
• C. \( \dfrac{a - 2b}{a + 2b} \)
• D. \( \dfrac{2a + b}{2a - b} \)

Hint:

Use implicit differentiation carefully when both variables are involved in a product. Always simplify expressions by canceling common terms.

Answer 1

The Correct Option is B

We are given: \[ (a + \sqrt{2}b \cos x)(a - \sqrt{2}b \cos y) = a^2 - b^2 \] Differentiate both sides implicitly with respect to \( x \): Let \( u = (a + \sqrt{2}b \cos x),
v = (a - \sqrt{2}b \cos y) \), so that: \[ uv = a^2 - b^2 \Rightarrow \frac{d}{dx}(uv) = 0 \Rightarrow \frac{du}{dx}v + u \frac{dv}{dx} = 0 \] Now compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \): \[ \frac{du}{dx} = -\sqrt{2}b \sin x,
\frac{dv}{dx} = \sqrt{2}b \sin y . \frac{dy}{dx} \] Substitute into derivative: \[ (-\sqrt{2}b \sin x)(a - \sqrt{2}b \cos y) + (a + \sqrt{2}b \cos x)(\sqrt{2}b \sin y . \frac{dy}{dx}) = 0 \] \[ \Rightarrow -\sqrt{2}b \sin x(a - \sqrt{2}b \cos y) + \sqrt{2}b \sin y(a + \sqrt{2}b \cos x) . \frac{dy}{dx} = 0 \] Solve for \( \dfrac{dy}{dx} \): \[ \Rightarrow \frac{dy}{dx} = \frac{\sqrt{2}b \sin x(a - \sqrt{2}b \cos y)}{\sqrt{2}b \sin y(a + \sqrt{2}b \cos x)} \] Cancel \( \sqrt{2}b \) in numerator and denominator: \[ \frac{dy}{dx} = \frac{\sin x(a - \sqrt{2}b \cos y)}{\sin y(a + \sqrt{2}b \cos x)} \] Now plug in \( x = \frac{\pi}{4}, y = \frac{\pi}{4} \). Then: \[ \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}},
\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}} \] \[ \Rightarrow \frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}}(a - \sqrt{2}b . \frac{1}{\sqrt{2}})}{\frac{1}{\sqrt{2}}(a + \sqrt{2}b . \frac{1}{\sqrt{2}})} = \frac{a - b}{a + b} \]