Question

If \(\sqrt{y+x}+\sqrt{y-x} = c\) and \(\frac{dy}{dx} = K-\sqrt{\frac{y^2}{x^2}-1}\) , then the value of \(K\) is

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

Hint:

Square the relation and differentiate implicitly, then simplify.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Differentiate \(\sqrt{y+x} + \sqrt{y-x} = c\) with respect to \(x\), then solve for \(\frac{dy}{dx}\).

Step 2: Differentiate:
\[ \frac{y' + 1}{2\sqrt{y+x}} + \frac{y' - 1}{2\sqrt{y-x}} = 0 \]
Multiply through: \((y'+1)\sqrt{y-x} + (y'-1)\sqrt{y+x} = 0\).
So \(y'(\sqrt{y-x} + \sqrt{y+x}) = \sqrt{y+x} - \sqrt{y-x}\).

Step 3: Simplify:
\[ y' = \frac{\sqrt{y+x} - \sqrt{y-x}}{\sqrt{y+x} + \sqrt{y-x}} \]
Multiply numerator and denominator by \(\sqrt{y+x} - \sqrt{y-x}\):
\[ y' = \frac{(y+x) + (y-x) - 2\sqrt{y^2-x^2}}{2x} = \frac{y - \sqrt{y^2-x^2}}{x} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1} \]

Step 4: Compare:
Matching with \(\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2}-1}\) gives \(K = \frac{y}{x}\).

Final Answer:
\(K = \frac{y}{x}\), option (D). \[ \boxed{\frac{y}{x}} \]