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Length of the subtangent at $(a, a)$ on the curve $y^{2}=\frac{x^{2}}{2 a+x}$ is equal to
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The correct answer is:
$\frac{18 a^{2}}{5}$
$y^{2}=\frac{x^{2}}{2 a+x}$
$2 y \frac{d y}{d x}=\frac{2 x}{2 a+x}-\frac{x^{2}}{(2 a+x)^{2}}$
At $(a, a),\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{18 a}$
Length of subtangent at $\left(x_{1}, y_{1}\right)=\frac{y_{1}}{\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}}$
$\therefore$ Length of subtangent at $(a, a)=\frac{a}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{18 a^{2}}{5}$
$2 y \frac{d y}{d x}=\frac{2 x}{2 a+x}-\frac{x^{2}}{(2 a+x)^{2}}$
At $(a, a),\left(\frac{d y}{d x}\right)_{(a, a)}=\frac{5}{18 a}$
Length of subtangent at $\left(x_{1}, y_{1}\right)=\frac{y_{1}}{\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}}$
$\therefore$ Length of subtangent at $(a, a)=\frac{a}{\left(\frac{d y}{d x}\right)_{(a, a)}}=\frac{18 a^{2}}{5}$
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