Question

The curve represented by the equation $\sqrt{px}+\sqrt{qy}=1$, where $p,q\in R$, $p,q>0$ is ?

• A. a circle
• B. a parabola
• C. an ellipse
• D. a hyperbola

Answer 1

The correct option is B a parabola

We have, $\sqrt{px}+\sqrt{qy}=1$
$\Rightarrow (\sqrt{px}+\sqrt{qy}{)}^2=1$
$\Rightarrow px+qy+2\sqrt{\left(pq\right)\left(xy\right)}=1$
$\Rightarrow (px+qy-1{)}^2=4(pq\left)\right(xy)$
$\Rightarrow p^2{x}^2-2\left(pq\right)\left(xy\right)+q^2{y}^2-2px-2qy+1=0$
On comparing this equation with the equation
$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, we get
$a=p^2,b=q^2,c=1,g=-p,$
$f=-q$ and $h=-pq$
$\Delta =abc+2fgh-a{f}^2-b{g}^2-c{h}^2$
$p^2{q}^2-2p^2{q}^2-p^2{q}^2-p^2{q}^2-p^2{q}^2$
$=-4p^2{q}^2\ne 0$
and, $h^2-ab=p^2{q}^2-p^2{q}^2=0$
Thus, we have $\Delta \ne 0$ and $h^2=ab$
Hence, the given curve is parabola.