If the pair of straight lines x2−y2+6x+4y+5=0 represents transverse and conjugate axes of the hyperbola and centre of hyberbola is (α,β), then value of β−α is
A
5
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B
5.0
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C
5.00
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Solution
x2−y2+6x+4y+5=0.....(1) ∵x2−y2=(x−y)(x+y)
Let the seperate equation of line be (x−y+l)(x+y+m)
In order to find value's of l and m, we have to equate the coefficients of x and y x2−xy+lx+xy−y2+ly+mx−my+lm=0 x2−y2+(l+m)x+(l−m)y+lm=0....(2)
Comparing (1) & (2), we get l+m=6 l−m=4 2l=10 ∴l=5 and m=1
So equations of axes are x−y+5=0 and x+y+1=0
Centre will be point of intersection of both the lines, so x=−3 and y=2 ∴ Centre is (−3,2)=(α,β) ⇒β−α=5