NIMCET 2008 Question Paper with solutions

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Question : 2 of 120

Marks: +1, -0

If

y=m x

bisects the angle between the lines

x^{2}(\tan ^{2} θ+\cos ^{2} θ)+2 x y \tan θ-y^{2} \sin ^{2} θ=0

when

θ= {π}/{3}

then the value of

√{3} m^{2}+4 m

1
$1/√3$
$√3$
$7√3$

Solution:

Equation of angle bisectors of

a x^{2}+2 h x y+b y^{2}=0

is

{x^{'}-y^{'}}/{a-b}= {x y}/{h}

⇒ Angle bisectors of

x^{2}(\tan ^{2} θ+\cos ^{2} θ+2 x y \tan θ-y^{2} \sin ^{2} θ=0)

{x^{2}-y^{2}}/{\tan ^{2} θ+\cos ^{2} θ+\sin ^{2} θ}=\;{x y}/{\tan θ}

⇒ {x^{2}-y^{2}}/{\sec ^{2} θ}=\;{x y}/{\tan θ} \;\; ( ∵ θ=π / 3)

⇒ {x^{2}-y^{2}}/{4}=\;{x y}/{√{3}}

. . .(i)
As

y=m x

satisfy Eq. (1), so

\;{x^{2}-m^{2} x^{2}}/{4}=\;{m x^{2}}/{√{3}}

\Rightarrow

\;{1-m^{2}}/{4}=\;{m}/{√{3}}

⇒ √{3}-√{3} m^{2}=4 m

⇒ √{3} m^{2}+4 m=√{3}

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