The conic represented by the equation 2xy+4x−6y+17=0 is
A
an ellipse
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B
a pair of straight line
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C
a parabola
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D
a hyperbola
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Solution
The correct option is D a hyperbola On compairing the given equation with ax2+by2+2hxy+2gx+2fy+c=0, we get a=b=0,c=17,g=2,f=−3,h=1 ∴Δ=abc+2fgh−af2−bg2−ch2=−29≠0 and h2−ab=1−0>0 Hence Δ≠0,h2>ab So the given equation represents a hyperbola.