Question

If \(\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}} = 6\), then \(\frac{dy}{dx} =\)

• A. \(\frac{x+17y}{17x-y}\)
• B. \(\frac{x-17y}{17x-y}\)
• C. \(\frac{x-17y}{17x+y}\)
• D. \(\frac{x+17y}{17x+y}\)

Hint:

Let t = sqrt(x/y) and turn the equation into a relation between x and y without roots.

Answer 1

The Correct Option is B

Step 1: Approach
Use homogeneity. Put $y=vx$ so the equation has only $v$.

Step 2: Find v
$\dfrac1{\sqrt v}+\sqrt v=6$, which fixes $v$ to a constant value. So the curve is a straight line $y=vx$ through the origin, and $\dfrac{dy}{dx}=v=\dfrac yx$.

Step 3: Match with the options
On the curve $x^2+y^2=34xy$. Then $x^2-17xy=17xy-y^2$, that is $x(x-17y)=y(17x-y)$, so $\dfrac yx=\dfrac{x-17y}{17x-y}$.

Step 4: Conclusion
$\dfrac{dy}{dx}=\dfrac{x-17y}{17x-y}$, option (B).

Final Answer:
After squaring, x squared + y squared = 34xy, and differentiating gives (x - 17y)/(17x - y), option (B). \[ \boxed{\frac{x-17y}{17x-y}} \]