The product of the perpendiculars from (1,1) to the pair of straight lines 2x^2 + 6xy + 3y^2 = 0 is

And:- 11/(37)^0.5

I have been trying this question 5 or 6 times now and still bot getting the answer... Can anyone please help me out ? Thanks in advance :)

Question

The product of the perpendiculars from (1,1) to the pair of straight lines 2x^2 + 6xy + 3y^2 = 0 is

And:- 11/(37)^0.5

I have been trying this question 5 or 6 times now and still bot getting the answer... Can anyone please help me out ? Thanks in advance :)

Badiuddin askIITians.ismu Expert · Answer

Dear Gautham let y=m 1 x and y=m 2 x are two lines and p 1 and p 2 are the length of perpendicular from (1,1) on these line respectively the m 1 +m 2 =-6/3 =-2 m 1 .m 2 =2/3 p 1 =(1*1 -1*m 1 )/&radic;(1+m 1 2 ) =(1-m 1 )/&radic;(1+m 1 2 ) similerly p 2 =(1*1 -1*m 1 )/&radic;(1+m 1 2 ) =(1-m 2 )/&radic;(1+m 2 2 ) so p1 .p2 ={(1-m 1 )/&radic;(1+m 1 2 )}*{(1-m 2 )/&radic;(1+m 2 2 )} ={1-(m 1+ m 2) + m 1 m 2}/&radic;{1+( m 1 m 2) 2 +( m 1 + m 2 ) 2 -2 m 1 m 2} ={1+2+2/3}/&radic;{1+ 4/9 +4 -4/3 } =11/&radic;37 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin

Gautham Warrier · Answer

Thanks so much :) A lot faster method than what I thought was the method :)

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The product of the perpendiculars from (1,1) to the pair of straight lines 2x^2 + 6xy + 3y^2 = 0 is And:- 11/(37)^0.5 I have been trying this question 5 or 6 times now and still bot getting the answer... Can anyone please help me out ? Thanks in advance :)

The product of the perpendiculars from (1,1) to the pair of straight lines 2x^2 + 6xy + 3y^2 = 0 is

And:- 11/(37)^0.5

I have been trying this question 5 or 6 times now and still bot getting the answer... Can anyone please help me out ? Thanks in advance :)

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