Question

The slope of tangent at any point (x, y) on a curve y = y(x) is \(\frac{x^2+y^2}{2xy},x>0\) If y(2) = 0, then a value of y(8) is
 

• A. 4√3
• B. -4√2
• C. 2√3
• D. -2√3

Answer 1

The Correct Option is A

Given:

\( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \).

• Step 1: Substitute \( y = vx \), where \( \frac{dy}{dx} = v + x \frac{dv}{dx} \):

  \( v + x \frac{dv}{dx} = \frac{x^2 + v^2x^2}{2vx^2} \).

  Simplify:

  \( v + x \frac{dv}{dx} = \frac{v^2 + 1}{2v} \).

• Step 2: Simplify and separate variables:

  \( x \cdot \frac{dv}{dx} = \frac{v^2 + 1 - 2v^2}{2v} = \frac{1 - v^2}{2v} \).

  Rearrange:

  \( \frac{2v \, dv}{1 - v^2} = \frac{dx}{x} \).

• Step 3: Integrate both sides:

  \( \int \frac{2v \, dv}{1 - v^2} = \int \frac{dx}{x} \).

  The left-hand side simplifies to:

  \( -\ln|1 - v^2| \).

  The right-hand side becomes:

  \( \ln|x| + C \).

  Combine:

  \( -\ln|1 - v^2| = \ln|x| + C \).

  Simplify:

  \( \ln|x(1 - v^2)| = C \).

  Back-substitute \( v = \frac{y}{x} \):

  \( \ln\left(\frac{x^2 - y^2}{x}\right) = C \).

  Simplify:

  \( x^2 - y^2 = cx \).

• Step 4: Apply the initial conditions \( x = 2, y = 0 \):

  \( 2^2 - 0^2 = c(2) \implies c = 2 \).

  The equation becomes:

  \( x^2 - y^2 = 2x \).

• Step 5: Find \( y(8) \) when \( x = 8 \):

  \( 8^2 - y^2 = 2(8) \).

  Simplify:

  \( 64 - y^2 = 16 \implies y^2 = 48 \implies y = \sqrt{48} = 4\sqrt{3} \).

Final Answer: \( y(8) = 4\sqrt{3} \).