Question

If $x^2+y^2=a^2$ and $k=\frac{1}{a},$ then $k$ is equal to

• A. $\frac{\left|y ''\right|}{\sqrt{1+y '}}$
• B. $\frac{\left|y ''\right|}{\sqrt{\left(1+y '^{2}\right)^{3}}}$
• C. $\frac{2y ''}{\sqrt{1+y '^{2}}}$
• D. $\frac{2y ''}{2\sqrt{\left(1+y '^{2}\right)}^{3}}$

Answer 1

Answer: (B)

$x^{2}+y^{2}=a^{2} \Rightarrow 2 x+2 y y^{\prime}=0 \Rightarrow y^{\prime}=-x / y$
$\Rightarrow y y^{\prime}+x=0$
$\Rightarrow y y^{\prime \prime}+y^{\prime 2}+1=0 \Rightarrow y=-\left(\frac{1+y^{\prime 2}}{y^{\prime \prime}}\right) \ldots(i)$
$\therefore k=\frac{1}{a}=\left|\frac{1}{\sqrt{x^{2}+y^{2}}}\right|=\left|\frac{1}{y \sqrt{1+\frac{x^{2}}{y^{2}}}}\right|=\left|\frac{1}{y \sqrt{1+y_{1}^{2}}}\right|\left[\because y^{\prime}=\frac{x}{y}\right]$
$=\left|\frac{-y^{\prime \prime}}{\left(1+y^{\prime 2}\right) \sqrt{1+y^{\prime 2}}}\right|=\frac{\left|y^{\prime \prime}\right|}{\left(1+y^{\prime 2}\right)^{3 / 2}}$