Question

If the tangent to the curve \(xy+ax+by = 0\) at \((1,1)\) makes an angle of \(tan^{-1}2\) with positive direction of the \(x\)-axis, then the value of \(\frac{ab}{a+b}\) is...

• A. \(1\)
• B. \(-1\)
• C. \(2\)
• D. \(-2\)

Hint:

Use the point to get one equation and the slope tan(angle) = 2 to get another.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The slope of the tangent at a point is \(\frac{dy}{dx}\). The tangent makes an angle \(\tan^{-1}2\) with the positive x-axis, so its slope is \(\tan(\tan^{-1}2) = 2\).

Step 2: Key Formula or Approach:
The curve passes through \((1,1)\), so \(1 + a + b = 0\). Differentiate \(xy + ax + by = 0\) implicitly.

Step 3: Detailed Explanation:
\(y + x\frac{dy}{dx} + a + b\frac{dy}{dx} = 0\), so \(\frac{dy}{dx} = -\frac{y + a}{x + b}\).
At \((1,1)\): \(-\frac{1 + a}{1 + b} = 2\), so \(1 + a = -2 - 2b\), which gives \(a + 2b = -3\).
From \(a + b = -1\): subtract to get \(b = -2\), and then \(a = 1\).
\[ \frac{ab}{a + b} = \frac{(1)(-2)}{-1} = 2 \]

Final Answer:
The value is \(2\), option (C). \[ \boxed{2} \]