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Question

The curve represented by the equation px+qy=1, where p,qR, p,q>0 is ?

A
a circle
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B
a parabola
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C
an ellipse
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D
a hyperbola
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Solution

The correct option is B a parabola
We have, px+qy=1
(px+qy)2=1
px+qy+2(pq)(xy)=1
(px+qy1)2=4(pq)(xy)
p2x22(pq)(xy)+q2y22px2qy+1=0
On comparing this equation with the equation
ax2+2hxy+by2+2gx+2fy+c=0, we get
a=p2,b=q2,c=1,g=p,
f=q and h=pq
Δ=abc+2fghaf2bg2ch2
p2q22p2q2p2q2p2q2p2q2
=4p2q20
and, h2ab=p2q2p2q2=0
Thus, we have Δ0 and h2=ab
Hence, the given curve is parabola.

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