$\therefore 2\left( {x - a} \right) + 2\left( {y - b} \right)\frac{{dy}}{{dx}} = 0$
$\Rightarrow 2\left( {x - a} \right) = - 2\left( {y - b} \right)\frac{{dy}}{{dx}}$
$\Rightarrow \frac{{dy}}{{dx}} = - \left( {\frac{{x - a}}{{y - b}}} \right)$ ……….(ii)
Again $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right).1 - \left( {x - a} \right)\frac{{dy}}{{dx}}} \right]}}{{{{\left( {y - b} \right)}^2}}}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right).1 - \left( {x - a} \right)\left( {\frac{{ - \left( {x - a} \right)}}{{y - b}}} \right)} \right]}}{{{{\left( {y - b} \right)}^2}}}$ [From eq. (ii)
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - \left[ {\left( {y - b} \right) + \left( {\frac{{{{\left( {x - a} \right)}^2}}}{{y - b}}} \right)} \right]}}{{{{\left( {y - b} \right)}^2}}}$
$= \frac{{ - \left[ {{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}} \right]}}{{{{\left( {y - b} \right)}^3}}}$
$= \frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}$ ……….(iii) [using (i)]
Putting values of $\frac{{dy}}{{dx}}$ and $\frac{{{d^2}y}}{{d{x^2}}}$ in the given expression,
$\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}$
$ = \frac{{{{\left[ {1 + {{\frac{{\left( {x - a} \right)}}{{{{\left( {y - b} \right)}^2}}}}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{ - {c^2}}}{{{{\left( {y - b} \right)}^3}}}}}$
$= \frac{{{{\left[ {{{\left( {y - b} \right)}^2} + {{\left( {x - a} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{{{\left( {y - b} \right)}^3}}} \times \frac{{{{\left( {y - b} \right)}^3}}}{{ - {c^2}}} = \frac{{{{\left( {{c^2}} \right)}^{\frac{3}{2}}}}}{{ - {c^2}}} = - c$
which is a constant and is independent of a and b.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.