Question

If $2x-ky+3=0,3x-y+1=0$ are conjugate lines with respect to $5x^2-6y^2=15$, then $k=$

• A. 6
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (A)

We know that,
If two lines $l_{1}x+m_{1}y+n_1=0$ and
$l_{2}x+m_{2}y+n_2=0$ are conjugate lines with respect to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $a^2{l}_1{l}_2-b^2{m}_1{m}_2=n_1{n}_2$
For hyperbola,
$5x^2-6y^2=15\Rightarrow \frac{x^2}{3}-\frac{y^2}{5/2}=1$
Here, $a^2=3,b^2=\frac{5}{2},l_1=2l_2=3$,
$m_1=-k,m_2=-1$
and $n_1=3,n_2=1$
$\therefore 3\times 2\times 3-\frac{5}{2}(-k)(-1)=3\times 1$
$\Rightarrow 18-\frac{5}{2}k=3\Rightarrow 15=\frac{5}{2}k\Rightarrow k=6$