Question

The value of $k$ so that the equation $12x^2-10y^2+11x-5y+k=0$ may represent a pair of straight lines is

• A. $k=\frac{91}{48}$
• B. $k=\frac{94}{43}$
  
• C. $k=\frac{83}{23}$
• D. None of these

Answer 1

The correct option is A $k=\frac{91}{48}$

$\Rightarrow$ The given equation is $12x^2-10y^2+11x-5y+k=0$

$\Rightarrow$ Comparing it with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$

$\Rightarrow$ $a=12,b=-10,h=o,g=\frac{11}{2},f=\frac{-5}{2},c=k$

$\Rightarrow$ The condition is $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$\Rightarrow$ $12\times (-10)\times k+2\times \left(\frac{-5}{2}\right)\times \frac{11}{2}\times 0-12\times (\frac{-5}{2}{)}^2-(-10)\times (\frac{11}{2}{)}^2-k\times (0{)}^2=0$

$\Rightarrow$ $-120k+0-75+\frac{605}{2}-0=0$

$\Rightarrow$ $\frac{-240k-150+605}{2}=0$

$\Rightarrow$ $-240k+455=0$

$\Rightarrow$ $k=\frac{455}{240}$

$\therefore$ $k=\frac{91}{48}$