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Question

If the lines ax2+2hxy+by2=0 be two sides of a parallelogram and the line lx+my=1 be one of its diagonal, then y (blhm)x (amhl)=?.

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Solution

Consider the diagram shown below.


Equation of OCy=m1x
Equation of OAy=m2x

Given:
Equation of AClx+my=1
Equation of the sides ax2+2hxy+by2=0

Therefore,
m1+m2=2hb
m1m2=ab

C is the intersection point of AC and OC. Therefore,
lx+m(m1x)=1
x=1l+mm1
y=m1x=m1l+mm1

Also, A is the intersection point of AC and OA. Therefore,
lx+m(m2x)=1
x=1l+mm2
y=m2x=m2l+mm2

Let P(α,β) be the mid-point of AC. Then,

α=12(1l+mm1+1l+mm2) ……. (1)

β=12(m1l+mm1+m2l+mm2) …… (2)


From equations (1) and (2), we have

αβ=2l+m(m1+m2)m1(l+mm2)+m2(l+mm1)

αβ=2l+m(2hb)l(2hb)+2m(ab)

αβ=blmhamhl


We know that, diagonals of a parallelogram bisect each other. So, P will also lie on OB. Two points on OB are (0,0) and P(α,β). Therefore, equation of OB is,
y=βαx
y=amhlblhmx
y(blhm)=x(amhl)
y(blhm)x(amhl)=0

Hence, this is the required answer.



1062103_1007736_ans_3f94e1c39beb48248dd88b177af2c11c.png

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