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 $k=$

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Aicharya · Accepted Answer

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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text {, then } a^2 l_1 l_2-b^2 m_1 m_2=n_1 n_2$ For hyperbola, $5 x^2-6 y^2=15 \Rightarrow \frac{x^2}{3}-\frac{y^2}{5 / 2}=1$ Here, $a^2=3, b^2=\frac{5}{2}, l_1=2, l_2=3$, and $m_1=-k, m_2=-1$ $\begin{array}{lc} \text {and } & n_1=3, n_2=1 \ \therefore & 3 \times 2 \times 3-\frac{5}{2}(-k)(-1)=3 \times 1 \ \Rightarrow & 18-\frac{5}{2} k=3 \Rightarrow 15=\frac{5}{2} k \Rightarrow k=6 \end{array}$

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 \(k=\)

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