Question

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

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

Answer 1

Answer: (B)

$x^2+y^2=a^2\Rightarrow 2x+2y{y}^{\prime }=0\Rightarrow y^{\prime }=-x/y$
$\Rightarrow y{y}^{\prime }+x=0$
$\Rightarrow y{y}^{\prime \prime }+y^{\prime 2}+1=0\Rightarrow y=-\left(\frac{1+y^{\prime 2}}{y^{\prime \prime }}\right)\dots (i)$
$\therefore k=\frac{1}{a}=\left|\frac{1}{\sqrt{x^2+y^2}}\right|=\left|\frac{1}{y\sqrt{1+\frac{x^2}{y^2}}}\right|=\left|\frac{1}{y\sqrt{1+y_1^2}}\right|\left[\because y^{\prime }=\frac{x}{y}\right]$
$=\left|\frac{-y^{\prime \prime }}{\left(1+y^{\prime 2}\right)\sqrt{1+y^{\prime 2}}}\right|=\frac{|y^{\prime \prime }|}{{\left(1+y^{\prime 2}\right)}^{3/2}}$