If $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0,$ then show that $\frac{{dy}}{{dx}} \cdot \frac{{dx}}{{dy}} = 1$

We have, $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ ……(i)

On differentiating both sides w.r.t. $x$, we get
$\frac{d}{{dx}}\left( {a{x^2}} \right) + \frac{d}{{dx}}(2hxy) + \frac{d}{{dx}}\left( {b{y^2}} \right) + \frac{d}{{dx}}(2gx) + \frac{d}{{dx}}(2fy) + \frac{d}{{dx}}(c) = 0$

$\Rightarrow$ $2ax + 2h\left( {x \cdot \frac{{dy}}{{dx}} + y \cdot 1} \right) + b \cdot 2y\frac{{dy}}{{dx}} + 2g + 2f\frac{{dy}}{{dx}} + 0 = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}}[2hx + 2by + 2f] = - 2ax - 2hy - 2g$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 2(ax + hy + g)}}{{2(hx + by + f)}}$

$= \frac{{ - (ax + hy + g)}}{{(hx + by + f)}}$ ……….(ii)

Now, differentiating Eq. (i) w.r.t. $y$, we get
$\frac{d}{{dy}}\left( {a{x^2}} \right) + \frac{d}{{dy}}(2hxy) + \frac{d}{{dy}}\left( {b{y^2}} \right) + \frac{d}{{dy}}(2gx) + \frac{d}{{dy}}(2fy) + \frac{d}{{dy}}(c) = 0$

$\Rightarrow$ $a \cdot 2x \cdot \frac{{dx}}{{dy}} + 2h \cdot \left( {x \cdot \frac{d}{{dy}}y + y \cdot \frac{d}{{dy}}x} \right) + b \cdot 2y + 2g \cdot \frac{{dx}}{{dy}} + 2f + 0 = 0$

$\Rightarrow$ $\frac{{dx}}{{dy}}[2ax + 2hy + 2g] = - 2hx - 2by - 2f$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \frac{{ - 2(hx + by + f)}}{{2(ax + hy + g)}} = \frac{{ - (hx + by + f)}}{{(ax + hy + g)}}$ …….(iii)

therefore,$\frac{{dy}}{{dx}} \cdot \frac{{dx}}{{dy}} = \frac{{ - (ax + hy + g)}}{{(hx + by + f)}} \cdot \frac{{ - (hx + by + f)}}{{(ax + hy + g)}}$ [using Eqs. (ii) and (iii)]

$= 1 = {\rm{RHS}}$

Hence proved.