Question

If $2 x-k y+3=0,3 x-y+1=0$ are conjugate lines with respect to $5 x^{2}-6 y^{2}=15$, 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,
$5\, x^{2}-6 \,y^{2}=15 \Rightarrow \frac{x^{2}}{3}-\frac{y^{2}}{5 / 2}=1$
Here, $a^{2}=3, b^{2}=\frac{5}{2}, l_{1}=2 l_{2}=3$,
$m_{1}=-k, m_{2}=-1$
and $n_{1}=3, n_{2}=1$
$\therefore 3 \times 2 \times 3-\frac{5}{2}(-k)(-1)=3 \times 1$
$\Rightarrow 18-\frac{5}{2} k=3 \Rightarrow 15=\frac{5}{2} k \Rightarrow k=6$