Question Number 39747 by Raj Singh last updated on 10/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18
$${y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}{a}−{x}}\:\:\:\frac{{dy}}{{dx}}=\frac{\left(\mathrm{2}{a}−{x}\right)\mathrm{2}{x}−{x}^{\mathrm{2}} \left(\mathrm{0}−\mathrm{1}\right)}{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{4}{ax}−\mathrm{2}{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} }=\frac{\mathrm{4}{ax}−{x}^{\mathrm{2}} }{\left(\mathrm{2}{a}−{x}\right)^{\mathrm{2}} } \\ $$$${slope}\:{of}\:{tangent}\:{m}=\frac{\mathrm{4}{a}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{3} \\ $$$${eqn}\:{tangent}\:\left({y}−{a}\right)=\mathrm{3}\left({x}−{a}\right) \\ $$$${eqnnormal}\:\:{y}−{a}=\frac{−\mathrm{1}}{\mathrm{3}}\left({x}−{a}\right) \\ $$$${tangent}\:{cut}\:{x}\:{axis}\:{at}\:\left(\frac{\mathrm{2}{a}}{\mathrm{3}},\mathrm{0}\right) \\ $$$${normal}\:{cut}\:{x}\:{axis}\:\:\left(\frac{\mathrm{4}{a}}{},\mathrm{0}\right) \\ $$$${triangle}\:{formed}\:{by}\:\left({a},{a}\right)\:,\left(\mathrm{4}{a},\mathrm{0}\right)\:,\left(\frac{\mathrm{2}{a}}{\mathrm{3}},\mathrm{0}\right) \\ $$$$ \\ $$$${area}=\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{4}{a}−\frac{\mathrm{2}{a}}{\mathrm{3}}\right)×{a} \\ $$$$=\frac{\mathrm{5}{a}^{\mathrm{2}} }{\mathrm{3}}\:\:\left[{pls}\:{check}…\right. \\ $$