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If $x+y-1=0$ and $2 x-y+1=0$ are conjugate lines with respect to a circle $x^2+y^2-4 x+2 f y-1=0$, then $f=$
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-2 or 0
Conjugate lines are : $x+y-1=0,2 x-y+1=0$
$\mathrm{l}_1=1, \mathrm{~m}_1=1, \mathrm{n}_1=-1, \mathrm{l}_2=2, \mathrm{~m}_2=-1, \mathrm{n}_2=1$
$\begin{aligned} & \therefore \quad g=-2, c=-1 \\ & \text { Radius, } R=\sqrt{f^2+4+1}=\sqrt{f^2+5}\end{aligned}$
Now, the condition for the conjugate lines :
$\begin{aligned} & \mathrm{R}^2\left(\mathrm{l}_1 \mathrm{l}_2+\mathrm{m}_1 \mathrm{~m}_2\right)=\left(\mathrm{l}_1 \mathrm{~g}+\mathrm{m}_1 \mathrm{f}-\mathrm{n}_1\right)\left(\mathrm{l}_2 \mathrm{~g}+\mathrm{m}_2 \mathrm{f}-\mathrm{n}_2\right) \\ & \left(\mathrm{f}^2+5\right)(1.2+1(-1))=(1(-2)+1 \mathrm{f}+1)(2(-2)+(-1) \\ & \mathrm{f}-1) \\ & \Rightarrow \mathrm{f}^2+5=(\mathrm{f}-1)(-\mathrm{f}-5) \\ & \Rightarrow \mathrm{f}^2+2 \mathrm{f}=0 \\ & \Rightarrow \mathrm{f}=0 \text { or }-2 .\end{aligned}$
$\mathrm{l}_1=1, \mathrm{~m}_1=1, \mathrm{n}_1=-1, \mathrm{l}_2=2, \mathrm{~m}_2=-1, \mathrm{n}_2=1$
$\begin{aligned} & \therefore \quad g=-2, c=-1 \\ & \text { Radius, } R=\sqrt{f^2+4+1}=\sqrt{f^2+5}\end{aligned}$
Now, the condition for the conjugate lines :
$\begin{aligned} & \mathrm{R}^2\left(\mathrm{l}_1 \mathrm{l}_2+\mathrm{m}_1 \mathrm{~m}_2\right)=\left(\mathrm{l}_1 \mathrm{~g}+\mathrm{m}_1 \mathrm{f}-\mathrm{n}_1\right)\left(\mathrm{l}_2 \mathrm{~g}+\mathrm{m}_2 \mathrm{f}-\mathrm{n}_2\right) \\ & \left(\mathrm{f}^2+5\right)(1.2+1(-1))=(1(-2)+1 \mathrm{f}+1)(2(-2)+(-1) \\ & \mathrm{f}-1) \\ & \Rightarrow \mathrm{f}^2+5=(\mathrm{f}-1)(-\mathrm{f}-5) \\ & \Rightarrow \mathrm{f}^2+2 \mathrm{f}=0 \\ & \Rightarrow \mathrm{f}=0 \text { or }-2 .\end{aligned}$
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