Question

If \( (a^2 - 1)x + ay + (3 - a) = 0 \) is a normal to the curve \( xy = 1 \), then the interval in which ‘a’ lies is:

• A. \( [-1, 1] \cup [2, \infty) \)
• B. \( (-\infty, -1] \cup (0, 1] \)
• C. \( (-1, 1) \cup (1, \infty) \)
• D. \( (1, \infty) \)

Hint:

Use derivative-based slope comparison to determine when a line is tangent or normal to a curve.

Answer 1

The Correct Option is B

Given that line \( (a^2 - 1)x + ay + (3 - a) = 0 \) is normal to curve \( xy = 1 \). Find slope of the normal to the curve \( xy = 1 \). The curve’s derivative using implicit differentiation: \[ \frac{d}{dx}(xy) = \frac{d}{dx}(1) \Rightarrow x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \] So, slope of normal = \( \frac{x}{y} \) Compare with line in general form: \[ (a^2 - 1)x + ay + (3 - a) = 0 \Rightarrow \text{slope} = -\frac{a^2 - 1}{a} \] Equating: \[ \frac{x}{y} = -\frac{a^2 - 1}{a} \Rightarrow a x + (a^2 - 1)y = 0 \] Solve this system with \( xy = 1 \) to find feasible values for \( a \). From the solution, we find the values of \( a \) that make the line normal to the curve lie in the interval: \[ a \in (-\infty, -1] \cup (0, 1] \]