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Q.
The conic represented by the equation $\sqrt{ax} + \sqrt{by} = 1$ is
Conic Sections
Solution:
The given conic is $\sqrt{ax} + \sqrt{by} = 1$
Squaring both sides, $ax + by + 2 \sqrt{abxy} = 1$
or $ax + by -1 = - 2\sqrt{abxy}$
Squaring again, $\left(ax+ by- 1\right)^{2} = 4abxy$
or $a^{2}x^{2} - 2abxy + b^{2}+y^{2} - 2ax-2by+1 = 0\quad\dots \left(1\right)$
Comparing the equation $\left(1\right)$ with the equation
$Ax^{2} + 2Hxy +By^{2} +2Gx+2Fy+C = 0$
$\therefore \quad A = a^{2} , H = -ab, B = b^{2} , G = - a, F = -b, C = 1$
Then, $\Delta = ABC +2FGH -AF^{2}-BG^{2} - CH^{2}$
$= a^{2}b^{2} -2a^{2}b^{2}-a^{2}b^{2}-a^{2}b^{2}-a^{2}b^{2} = -4a^{2}b^{2} \ne 0$
and $H^{2} = a^{2}b^{2} = AB$
So we have $\Delta \ne 0$ and $H^{2} - AB = 0$. Hence the given equation represent a parabola.