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
- 4√3
- -4√2
- 2√3
- -2√3
The Correct Option is A
Solution and Explanation
Given:
\( \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \).
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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:\( v + x \frac{dv}{dx} = \frac{v^2 + 1}{2v} \).
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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:\( x^2 - y^2 = cx \).
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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: \( y(8) = 4\sqrt{3} \).
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