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If ax^2 +2hxy+by^2 +2gx + 2fy+c=0 & ax^2 + 2hxy + by^2 -2gx - 2fy +c=0 each represents pair of straight lines. Prove that the area of parallelogram enclosed by them is (2|c|)/((h^2 -ab) ^(1/2))

If ax^2 +2hxy+by^2 +2gx + 2fy+c=0 & ax^2 + 2hxy + by^2 -2gx - 2fy +c=0 each represents pair of straight lines. Prove that the area of parallelogram enclosed by them is (2|c|)/((h^2 -ab) ^(1/2))

Grade:12

2 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
ax2+2hxy+by2+2gx+2fy+c=0 is a couple of straight lines. It must be of the frame (lx+my+n)(px+qy+r)=0, where a=lp, h=12(lq+mp), b=mq, ..., c=nr (on looking at coefficients). Thus ax2+2hxy+by2−2gx−2fy+c=0 is then given by (lx+my−n)(px+qy−r)=0. 
Discriminant Δ = stomach muscle – h2 = - (lq – mp)2/4. 
The line lx+my+n=0 meets the lines px+qy±r=0 at the focuses (±(mr – nq)/(lq – mp), ±(lr – np)/(lq – mp)). The separation between two focuses = 2r √(l2 + m2)/(lq – mp) = Length of the side of the parallelogram = l1. 
The opposite separation between the parallel lines lx+my±n=0 = 2n/√(l2 + m2) = Distance between inverse sides of the parallelogram = l2. 
In this way territory = l1l2 = |2nr/(1/2 x (lq – mp))| = 2|c|/√(- Δ).
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)
Supriya Bhowmik
26 Points
one year ago
show that one of the bisectors of the angles between the pair of lines ax^ overline 2 +2hxy+by^ 2 =0 will pass through the point of intersection of two lines a * x ^ 2 + 2hxy + b * y ^ 2 + 2gx + 2fy + c = 0 ifh(g ^ 2 - f ^ 2) = fg(a - b)

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