If the lines $2 x-y+3=0$ and $4 x+k y+3=0$ are conjugate with respect to the ellipse $5 x^2+6 y^2-15=0$, then $k$ equals

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Aicharya · Accepted Answer

$2 x-y+3=0$ and $4 x+k y+3=0$ are conjugate with respect to ellipse $5 x^2+6 y^2-15=0$ We know that $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ are conjugate with $\frac{x^2}{a^2}+\frac{y^2}{b}=1$ Then, $a^2 a_1 a_2+b^2 b_1 b_2=c_1 c_2$ $	herefore$ Here, $a_1=2, b_1=-1, c_1=3$ $\begin{aligned} & a_2=4, b_2=k, c_2=3 \ & a^2=\frac{15}{5}, b^2=\frac{15}{6}\end{aligned}$ $	herefore \frac{15}{5}(2)(4)-\left(\frac{15}{6}ight)(k)=9$ $\Rightarrow \quad k=6$

If the lines $2 x-y+3=0$ and $4 x+k y+3=0$ are conjugate with respect to the ellipse $5 x^2+6 y^2-15=0$,…

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