MHT CET · Maths · Quadratic Equation
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 \(\left|m_{1}-m_{2}\right|=\)
- A 1
- B 2
- C 4
- D 3
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
\(\left(\operatorname{scc}^{2} \theta-\sin ^{2} \theta\right) x^{2}-2 \tan \theta x y+\sin ^{2} \theta y^{2}=0\)
\(\sqrt{a x^{2}+2 b x y+b y^{2}}=0\)
\(\left|m_{1}-m_{2}\right|=?\)
\(\left|m_{1}-m_{2}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}\)
\(m_{1}+m_{2}=\frac{-2 h}{b}\)
\(m_{1} m_{2}=\frac{a}{b}\)
\(m_{1}-m_{2} \mid=\sqrt{\frac{4 \tan ^{2} \theta-4\left(\sec ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}\)
\(=2\)
\(\sqrt{a x^{2}+2 b x y+b y^{2}}=0\)
\(\left|m_{1}-m_{2}\right|=?\)
\(\left|m_{1}-m_{2}\right|=\sqrt{\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}}\)
\(m_{1}+m_{2}=\frac{-2 h}{b}\)
\(m_{1} m_{2}=\frac{a}{b}\)
\(m_{1}-m_{2} \mid=\sqrt{\frac{4 \tan ^{2} \theta-4\left(\sec ^{2} \theta-\sin ^{2} \theta\right)}{\sin ^{2} \theta}}\)
\(=2\)
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