Question

If $x+y-1=0$ and $2x - y + 1=0$ are conjugate lines with respect to the circle $x^2 + y^2 - 4x + 2fy - 1 = 0$, then $f=$

• A. $-1$ or $3$
• B. $1$ or $2$
• C. $-2$ or $0$
• D. $-1$ or $2$

Hint:

Conjugate Lines.
For a line to be conjugate to another, its pole must lie on that line. Use polar equation substitution and equate to line to find unknowns.

Answer 1

The Correct Option is C

Two lines are conjugate if the pole of one lies on the other. Let $L_1: x + y - 1 = 0$ and the pole be $(x_1, y_1)$. Polar of $(x_1, y_1)$ is: \[ (x_1 - 2)x + (y_1 + f)y + (-2x_1 + fy_1 - 1) = 0 \] Matching with $x + y - 1 = 0$, compare coefficients: \[ \text{(1)}\ x_1 - 2 = 1 \Rightarrow x_1 = 3, \quad \text{(2)}\ y_1 + f = 1 \Rightarrow y_1 = 1 - f \] Substitute into constant term: \[ -2(3) + f(1 - f) - 1 = -6 + f - f^2 - 1 = -7 + f - f^2 \] Set equal to $-1$: \[ -7 + f - f^2 = -1 \Rightarrow f^2 - f - 6 = 0 \Rightarrow f = 3 \text{ or } -2 \]