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Question

If the pair of straight lines x2y2+6x+4y+5=0 represents transverse and conjugate axes of the hyperbola and centre of hyberbola is (α,β), then value of βα is

A
5.00
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B
5
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C
5.0
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Solution

x2y2+6x+4y+5=0 .....(1)
x2y2=(xy)(x+y)
Let the seperate equation of line be (xy+l)(x+y+m)
In order to find value's of l and m, we have to equate the coefficients of x and y
x2xy+lx+xyy2+ly+mxmy+lm=0
x2y2+(l+m)x+(lm)y+lm=0 ....(2)
Comparing (1) & (2), we get
l+m=6
lm=4
2l=10
l=5 and m=1
So equations of axes are xy+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)
βα=5

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