Find the value of $k$ , if the lines $x + y + 2 = 0$ and $x - 2 y + k = 0$ are conjugate with respect to parabola $y^{2} + 4 x - 2 y - 3 = 0$ .

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Solution

$y^{2} + 4 x - 2 y - 3 = 0$
$y^{2} - 2 y + 1 + 4 x - 4 = 0 \Rightarrow {(y - 1)}^{2} = - 4 (x - 1)$
Conjugate lines are $x + y + 2 = 0, x - 2 y + k = 0$
$(x - 1) + (y - 1) + 4 = 0, (x - 1) - 2 (y - 1) + k - 1 = 0$
Let $y - 1 = Y, x - 1 = X$
So parabola is $Y^{2} = 4 \times (- 1) X$ and lines are
$X + Y + 4 = 0, X - 2 Y + k - 1 = 0$
for $l_{1} x + m_{1} y + n_{1} = 0, l_{2} x + m_{2} y + n_{2} = 0$ to be conjugate wrt. parabola $y^{2} = 4 a x$ , condition is $l_{1} n_{2} + l_{2} n_{1} = 2 a m_{1} m_{1}$
So for $X + Y + 4 = 0, X - 2 Y + (k - 1) = 0$ to be conjugate wrt $y^{2} = 4 (- 1) x$
$1 \times (k - 1) + 1 \times 4 = 2 \times (- 1) \times 1 \times - 2$
$\Rightarrow k - 1 + 4 = 4$
$\Rightarrow k = 1$

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Find the value of k, if the lines x+y+2=0 and x−2y+k=0 are conjugate with respect to parabola y2+4x−2y−3=0.

Find the value of $k$ , if the lines $x + y + 2 = 0$ and $x - 2 y + k = 0$ are conjugate with respect to parabola $y^{2} + 4 x - 2 y - 3 = 0$ .