If $m_{1}$ and $m_{2}$ are slopes of the lines represented by $\left(\sec ^{2} heta-\sin ^{2} heta ight) x^{2}-2 an heta x y+\sin ^{2} heta y^{2}=0$, then $\left|m_{1}-m_{2} ight|=$

Question

If $m_{1}$ and $m_{2}$ are slopes of the lines represented by $\left(\sec ^{2} 	heta-\sin ^{2} 	hetaight) x^{2}-2 	an 	heta x y+\sin ^{2} 	heta y^{2}=0$, then $\left|m_{1}-m_{2}ight|=$

Aicharya · Accepted Answer

$\left(\operatorname{scc}^{2} 	heta-\sin ^{2} 	hetaight) x^{2}-2 	an 	heta x y+\sin ^{2} 	heta y^{2}=0$ $\sqrt{a x^{2}+2 b x y+b y^{2}}=0$ $\left|m_{1}-m_{2}ight|=?$ $\left|m_{1}-m_{2}ight|=\sqrt{\left(m_{1}+m_{2}ight)^{2}-4 m_{1} m_{2}}$ $m_{1}+m_{2}=\frac{-2 h}{b}$ $m_{1} m_{2}=\frac{a}{b}$ $m_{1}-m_{2} \mid=\sqrt{\frac{4 	an ^{2} 	heta-4\left(\sec ^{2} 	heta-\sin ^{2} 	hetaight)}{\sin ^{2} 	heta}}$ $=2$

If $m_{1}$ and $m_{2}$ are slopes of the lines represented by $\left(\sec ^{2} \theta-\sin ^{2}…

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