Question

If the lines $2x+3y+12=0$ and $x-y+4k=0$ are conjugate with respect to the parabola $y^2=8x$, then the value of $k$ is:

• A. $-3$
• B. $3$
• C. $2$
• D. $-2$

Answer 1

Answer: (A)

$-3$

Given Parabola: $y^2=8x................\left(i\right)(a=2)$

Two conjugate lines: $L_1:2x+3y+12=0$

$L_2:x-y+4k=0$

Pole of $L_1=>(\frac{n}{l},\frac{-2am}{l}):P_1(\frac{12}{2},\frac{-2\left(2\right)\left(3\right)}{2})$

$=>P_1:(6,-6)$

As the lines are conjugate, $P_1$ must lie on $L_2$ $=>6-(-6)+4k=0$

$=>k=\frac{-12}{4}$

$=>k=-3$.