Question

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

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

Hint:

Differentiate the relation implicitly with respect to \(y\) (treat \(x\) as a function of \(y\)) and substitute \(x=y=\pi/4\).

Answer 1

The Correct Option is C

Check: at \(x=y=\pi/4\), \(\sqrt{2}\cos(\pi/4)=1\), so LHS \(=(a+b)(a-b)=a^2-b^2\), so the point lies on the curve.
Differentiate both sides with respect to \(y\):
\((-\sqrt{2}\,b\sin x\,\frac{dx}{dy})(a-\sqrt{2}\,b\cos y)+(a+\sqrt{2}\,b\cos x)(\sqrt{2}\,b\sin y)=0\).
At \((\pi/4,\pi/4)\): \(\sqrt{2}\sin(\pi/4)=1\), so \(-b(a-b)\frac{dx}{dy}+b(a+b)=0\).
Hence \(\frac{dx}{dy}=\frac{a+b}{a-b}\).
Answer: (3) \(\frac{a+b}{a-b}\)