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hello, i am looking for an answer to the question in maths which is related to the chapter pair of staight lines. the question is the slope of one line of 3x square + 2hxy + 8y square is double the slope of the other line then h square =
He let the slopes of two lines be m1 and m2. Therefore given is 2m1=m2.In the given equation there are two straight lines. Buy some derivation you get m1 + m2 = -2h/b and m1m2=a/b. Here the equation is of the form ax2 + 2hxy + by2. which gives a=3,b=8.Solving: m1+m2 = -2h/8 , but 2m1=m2.Therefore 3m1= -2h/8m1=-2h/24. (Eq-1).And m1m2=a/b,Therefore m1 (2m1)=3/82ml2=3/8m1= root3/4. (Eq-2).Equating 1 and 2 , -2h/24=root3/4-2h=6 (root3)h= (-3(root3))h2=(-3(root3))2h2=9×3=27For the proof related to those formulas please refer your notes or coaching material but I have solved you the problem....HOPE THIS HELPED YOU....
Are you the candidate of BASE BANGALORE. And if yes please refer to that material it's good even i am a candidate of that coaching institute.....
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