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Question

Mathematics Question on Application of derivatives

Lengthof the subtangent at (a,a)(a, a) on the curve y2=x22a+xy^2 = \frac{x^2}{2a +x} is equal to

A

185\frac{18}{5}

B

18a5\frac{18a}{5}

C

18a25\frac{18a^2}{5}

D

−18a25 - \frac{18a^2}{5}

Answer

18a25\frac{18a^2}{5}

Explanation

Solution

Given, y2=x22a+xy^{2} = \frac{x^{2}}{2a+x}
⇒2ydydx=(2a+x)(2x)−x2(1)(2a+x2)\Rightarrow 2y \frac{dy}{dx} = \frac{\left(2a+x\right)\left(2x\right)-x^{2} \left(1\right)}{\left(2a+x^{2}\right)}
⇒[dydx](a,a)=(2a+a)(2a)−a22×a(2a+a)2\Rightarrow \left[\frac{dy}{dx}\right]_{\left(a,a\right)} = \frac{\left(2a+a\right)\left(2a\right)-a^{2}}{2\times a\left(2a+a\right)^{2}}
=(3a)(2a)−a22a(3a)2= \frac{\left(3a\right)\left(2a\right)-a^{2}}{2a\left(3a\right)^{2}}
=6a2−a22a×9a2=5a218a3= \frac{6a^{2} -a^{2}}{2a \times9a^{2}} = \frac{5a^{2}}{18a^{3}}
length of subtangent at (a, a)
=y(dydx)(a,a)=a518a=a×18a5=18a25= \frac{y}{\left(\frac{dy}{dx}\right)_{\left(a,a\right)}}= \frac{a}{\frac{5}{18a}} = a\times \frac{18a}{5}=\frac{18a^{2}}{5}