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Question 126 of 139

Q.Find the angle between the lines represented by 3x2+4xy−3y2=03x^2+4xy-3y^2=0

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2020Subjective· 2mImportance★★★★★
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Use tan⁡θ=∣2h2−aba+b∣\tan\theta = \left|\dfrac{2\sqrt{h^2-ab}}{a+b}\right|; here a+b=0a+b=0.

Comparing 3x2+4xy−3y2=03x^2+4xy-3y^2=0 with ax2+2hxy+by2=0ax^2+2hxy+by^2=0: a=3, 2h=4⇒h=2, b=−3a=3,\ 2h=4 \Rightarrow h=2,\ b=-3.

a+b=3+(−3)=0a+b = 3+(-3) = 0

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