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If $m_{1}$ and $m_{2}$ are slopes of the lines represented by
$\left(\sec ^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0$, then $\left|m_{1}-m_{2}\right|=$
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$\left(\sec ^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0$, then $\left|m_{1}-m_{2}\right|=$
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Verified Answer
The correct answer is:
2
$\left(\operatorname{scc}^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0$
$\sqrt{a x^{2}+2 b x y+b y^{2}}=0$
$\left|m_{1}-m_{2}\right|=?$
$\left|m_{1}-m_{2}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$m_{1}+m_{2}=\frac{-2 h}{b}$
$m_{1} m_{2}=\frac{a}{b}$
$m_{1}-m_{2} \mid=\sqrt{\frac{4 \tan ^{2} \theta-4\left(\sec ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}$
$=2$
$\sqrt{a x^{2}+2 b x y+b y^{2}}=0$
$\left|m_{1}-m_{2}\right|=?$
$\left|m_{1}-m_{2}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}$
$m_{1}+m_{2}=\frac{-2 h}{b}$
$m_{1} m_{2}=\frac{a}{b}$
$m_{1}-m_{2} \mid=\sqrt{\frac{4 \tan ^{2} \theta-4\left(\sec ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}$
$=2$
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