Question

The slope of normal at any point \((x, y), x > 0, y > 0\) on the curve \(y = y(x)\) is given by \(\frac{x^2}{xy-x^2y2^-1}\) If the curve passes through the point (1, 1), then \(e · y(e)\) is equal to

• A. \(\frac{1-tan(1)}{1+tan(1)}\)
• B. \(tan(1)\)
• C. \(1\)
• D. \(\frac{1+tan(1)}{1-tan(1)}\)

Answer 1

The Correct Option is D

\(∵ -\frac{ dx}{dy} = \frac{x^2}{xy-x^2y^2-1}\)

\(\frac{dy}{dx} = \frac{x^2y^2-xy+1}{x^2}\)

Assuming \(xy = v ⇒ y+x \frac{dy}{dx} = \frac{dv}{dx}\)

\(\frac{dv}{dx}-y = \frac{(v^2+v+1)y}{v}\)

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

\(∵ y(1) = 1 ⇒ tan^{–1} (xy) = lnx + tan^{–1}(1)\)

Put \(x\) = \(e\) and \(y\) = \(y(e)\) we get

\(tan^{–1} (e · y(e)) = 1 + tan^{–1} 1\).

\(tan^{–1} (e · y(e)) – tan^{–1} 1 = 1\)

\(∴ e(y(e)) = \frac{1+tan(1)}{1-tan(1)}\)

Hence, the correct option is (D): \(\frac{1+tan(1)}{1-tan(1)}\)