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Question 114 of 129

Q.Find the separate equation of the pair of straight lines 3x2+2xy−y2=03x^2 + 2xy - y^2 = 0.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 2mImportance★★★★★
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Factoring 3x2+2xy−y23x^2+2xy-y^2 gives (3x−y)(x+y)(3x-y)(x+y), so the two lines are 3x−y=03x-y=0 and x+y=0x+y=0.

Treat 3x2+2xy−y2=03x^2+2xy-y^2=0 as a quadratic in xx: 3x2+(2y)x−y2=03x^2+(2y)x-y^2=0.

Using the quadratic formula (coefficients 3,2y,−y23, 2y, -y^2):

x=−2y±(2y)2−4(3)(−y2)2(3)=−2y±4y2+12y26=−2y±4y6x = \frac{-2y\pm\sqrt{(2y)^2-4(3)(-y^2)}}{2(3)} = \frac{-2y\pm\sqrt{4y^2+12y^2}}{6} = \frac{-2y\pm4y}{6}

This gives x=2y6=y3x=\dfrac{2y}{6}=\dfrac{y}{3} or x=−6y6=−yx=\dfrac{-6y}{6}=-y.

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