Question

If the tangent to the curve \( xy + ax + by = 0 \) at \( (1, 1) \) is inclined at an angle \( \tan^{-1}{2} \) with the X-axis, then

• A. \( a = 1, b = 2 \)
• B. \( a = 1, b = -2 \)
• C. \( a = -1, b = 2 \)
• D. \( a = -1, b = -2 \)

Hint:

When finding the slope of a curve at a given point, differentiate implicitly and use the point's coordinates to substitute values.

Answer 1

The Correct Option is B

Given the curve \( xy + ax + by = 0 \), we can differentiate it implicitly to find the slope of the tangent. 1. Differentiate \( xy + ax + by = 0 \) with respect to \( x \): \[ y + x\frac{dy}{dx} + a + b\frac{dy}{dx} = 0 \] Rearranging: \[ \frac{dy}{dx}(x + b) = -y - a \] Substituting \( x = 1 \) and \( y = 1 \) at point \( (1, 1) \): \[ \frac{dy}{dx}(1 + b) = -1 - a \] This gives us the slope of the tangent at \( (1, 1) \) as \( \frac{dy}{dx} = \frac{-1 - a}{1 + b} \). 2. The slope of the tangent is also \( \tan^{-1}{2} \), which means the slope is 2. Thus, \[ \frac{-1 - a}{1 + b} = 2 \] Solving for \( a \) and \( b \): \[ -1 - a = 2(1 + b) \quad \Rightarrow \quad -1 - a = 2 + 2b \quad \Rightarrow \quad a = -3 - 2b \] Substitute \( a = 1 \) and \( b = -2 \).