Question

If $x^2+y^2=a^2$ and $k=1/a$, then $k$ is equal to ?

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

Answer 1

The correct option is A $\frac{y\prime \prime }{\sqrt{1+y\prime }}$

$x^2+y^2=a^2$

$2x+2y{y}_1=0$

$y_1=\frac{-x}{y}$\\ $y{y}_1+x=0$

$y{y}_2+{y_1}^2+1=0$

$y=-\frac{1+{y_1}^2}{y_2}⟹\left(1\right)$

$k=\frac{1}{a}=\left|\frac{1}{\sqrt{x^2+y^2}}\right|=\left|\frac{1}{y\sqrt{1+\frac{x^2}{y^2}}}\right|$

$=\frac{1}{y}\sqrt{1+{y_1}^2}$

$=\frac{y^{\prime \prime }}{(1+{y_1}^2)\sqrt{1+{y_1}^2}}$

$=\frac{\left|y^{\prime \prime }\right|}{{(1+{y_1}^2)}^{\frac{3}{2}}}$