If the lines represented by the equation 2x2 - 3xy - y2 - 0 make angles - and - with x - axis- then cot2 - - cot2 - -

M_1 = tanα and m_2 = tanβ ⇒ cot α = 1/m_1 and cot β = 1/m_2 Hence, cot^2 α + cot^2 β = 1/m_1^2 + 1/m_2^2 = m_1^2 + m_2^2/(m_1m_2)^2 = (m_1 + m_2)^2 2m_1m_2 ...

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Byju's Answer

Standard XII

Mathematics

Pair of Lines Passing Though Origin

If the lines ...

Question

If the lines represented by the equation $2 x^{2} - 3 x y + y^{2} = 0$ make angles $α$ and $β$ with x - axis, then $c o t^{2} α + c o t^{2} β =$

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$\frac{3}{2}$

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$\frac{7}{4}$

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$\frac{5}{4}$

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Solution

The correct option is D
$\frac{5}{4}$

$m_{1} = t a n α a n d m_{2} = t a n β$

$\Rightarrow c o t α = \frac{1}{m_{1}}$ and $c o t β = \frac{1}{m_{2}}$

Hence, $c o t^{2} α + c o t^{2} β = \frac{1}{m_{1}^{2}} + \frac{1}{m_{2}^{2}} = \frac{m_{1}^{2} + m_{2}^{2}}{(m_{1} m_{2})^{2}}$

= $\frac{(m_{1} + m_{2})^{2} - 2 m_{1} m_{2}}{(m_{1} m_{2})^{2}} = \frac{(3)^{2} - 2 (2)}{(2)^{2}} = \frac{5}{4}$ .

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If the lines represented by the equation 2x2−3xy+y2=0 make angles α and β with x - axis, then cot2α+cot2β=

The correct option is D 54 m1=tanαandm2=tanβ ⇒cotα=1m1 and cotβ=1m2 Hence, cot2α+cot2β=1m21+1m22=m21+m22(m1m2)2 = (m1+m2)2−2m1m2(m1m2)2=(3)2−2(2)(2)2=54.

If the lines represented by the equation $2 x^{2} - 3 x y + y^{2} = 0$ make angles $α$ and $β$ with x - axis, then $c o t^{2} α + c o t^{2} β =$