Question Number 84236 by niroj last updated on 10/Mar/20
$$\:\:\mathrm{If}\:\boldsymbol{\mathrm{ax}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{hxy}}\:+\boldsymbol{\mathrm{by}}^{\mathrm{2}} =\:\mathrm{1},\:\mathrm{show}\:\mathrm{that} \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:=\:\frac{\boldsymbol{\mathrm{h}}^{\mathrm{2}} −\boldsymbol{\mathrm{ab}}}{\left(\boldsymbol{\mathrm{hx}}+\boldsymbol{\mathrm{by}}\right)^{\mathrm{3}} }. \\ $$
Answered by TANMAY PANACEA last updated on 10/Mar/20
$$\mathrm{2}{ax}+\mathrm{2}{hy}+\mathrm{2}{hx}.\frac{{dy}}{{dx}}+\mathrm{2}{by}.\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=−\left(\frac{{ax}+{hy}}{{hx}+{by}}\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(−\mathrm{1}\right)×\frac{\left({hx}+{by}\right)\left({a}+{h}\frac{{dy}}{{dx}}\right)−\left({ax}+{hy}\right)\left({h}+{b}×\frac{{dy}}{{dx}}\right)}{\left({hx}+{by}\right)^{\mathrm{2}} } \\ $$$$=\frac{\left({ax}+{hy}\right)\left\{\left({h}+{b}×\left(−\mathrm{1}\right)\left(\frac{{ax}+{hy}}{{hx}+{by}}\right)\right\}−\left({hx}+{by}\right)\left\{\left({a}+{h}×\left(−\mathrm{1}\right)×\frac{{ax}+{hy}}{{hx}+{by}}\right)\right\}\right.}{\left({hx}+{by}\right)^{\mathrm{2}\:} } \\ $$$$=\frac{\left({ax}+{hy}\right)\left({h}^{\mathrm{2}} {x}+{hby}−{abx}−{hby}\right)−\left({hx}+{by}\right)\left({ahx}+{aby}−{ahx}−{h}^{\mathrm{2}} {y}\right)}{\left({hx}+{by}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left({ax}+{hy}\right)\left({h}^{\mathrm{2}} {x}−{abx}\right)−\left({hx}+{by}\right)\left({aby}−{h}^{\mathrm{2}} {y}\right)}{\left({hx}+{by}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left({h}^{\mathrm{2}} −{ab}\right)\left({ax}^{\mathrm{2}} +{hxy}+{hxy}+{by}^{\mathrm{2}} \right)}{\left({hx}+{by}\right)^{\mathrm{3}\:} } \\ $$$$=\frac{\boldsymbol{{h}}^{\mathrm{2}} −\boldsymbol{{ab}}}{\left(\boldsymbol{{hx}}+\boldsymbol{{by}}\right)^{\mathrm{3}} }×\mathrm{1}=\frac{\boldsymbol{{h}}^{\mathrm{2}} −\boldsymbol{{ab}}}{\left(\boldsymbol{{hx}}+\boldsymbol{{by}}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$