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:

Answer 1

Hint:
We are going to solve this problem by using the definition of conjugate lines. Two lines are conjugate with respect to a parabola if each passes through the pole of the other. And two lines are conjugate if the pole of a line lies on another line and vice versa.

Complete step-by-step answer:
It is given that the lines \[2x + 3y + 12 = 0\] and \[x - 4y + 3k = 0\] are conjugate with respect to the parabola \[{y^2} = 8x\]
We are going to denote the line \[2x + 3y + 12 = 0\] by \[{L_1}\] and the line \[x - y + 4k = 0\] by \[{L_2}\]
Therefore, \[{L_1}:2x + 3y + 12 = 0\] and \[{L_2}:x - y + 4k = 0\]
Now we are going to apply the definition of conjugate lines with respect to a parabola.
Since the given lines are conjugate lines then the pole of a line lies on another line.
Hence we can say that the pole \[{P_1}\] of the line \[{L_1}:2x + 3y + 12 = 0\] lies on the other line \[{L_2}:x - y + 4k = 0\].
Now we are going to find the pole \[{P_1}\] of the line \[{L_1}:2x + 3y + 12 = 0\].
Now we are going to use the formula for finding the pole.
Coordinates of the pole of a given line \[lx + my + n = 0\] with respect to a parabola \[{y^2} = 4ax\] is given by,
\[\left( {\dfrac{n}{l},\dfrac{{ - 2am}}{l}} \right)\].
Comparing \[lx + my + n = 0\] and \[2x + 3y + 12 = 0\] we get, l=2, m=3 and n=12 and it is given that\[{y^2} = 8x = 4(2)x\]. Hence from this we find, a=2.
Now we are going to substitute the above values on \[\left( {\dfrac{n}{l},\dfrac{{ - 2am}}{l}} \right)\].
\[ \Rightarrow {P_1}:\left( {\dfrac{{12}}{2},\dfrac{{ - 2\left( 2 \right)\left( 3 \right)}}{2}} \right)\]
Now we simplify the above expression.
 \[ \Rightarrow {P_1}:\left( {6, - 6} \right)\]
Hence we finally found the pole of the line \[{L_1}:2x + 3y + 12 = 0\]
Now we already know that the pole \[{P_1}:\left( {6, - 6} \right)\] of the line \[{L_1}:2x + 3y + 12 = 0\] lies on the other line \[{L_2}:x - y + 4k = 0\].
We substitute 6 for x and -6 for y on \[{L_2}:x - y + 4k = 0\]
\[ \Rightarrow 6 - \left( { - 6} \right) + 4k = 0\]
\[
   \Rightarrow k = \dfrac{{ - 12}}{4} \\
   \Rightarrow k = - 3 \\
 \]
Hence, the value of k is -3

Note:
We must only substitute the pole of the line \[{L_1}\] on the other line \[{L_2}\]. We can’t substitute the pole of the line on the same line since by the definition of conjugate lines.