Question

If $x^2+y^2=R^2$ and $K=\frac{1}{R}$ then $K=$

• A. $\frac{y_1}{x\sqrt{1+y_1^2}}$
• B. $\frac{|y_2|}{\sqrt{(1+y_1^2{)}^3}}$
• C. $\frac{2|y_2|}{\sqrt{1+y_1^2}}$
• D. $\frac{3|y_2|}{\sqrt{(1+y_1^3{)}^3}}$

Answer 1

The correct option is A $\frac{y_1}{x\sqrt{1+y_1^2}}$

Differentiating $y^2+x^2=R^2$ we get

$2y{y}_1+2x=0\Rightarrow y_1=-\frac{x}{y}$

And again differentiating, we get

$y_2=-\left(\frac{y-x{y}_1}{y^2}\right)=-\left(\frac{y-x\times \left(\frac{-x}{y}\right)}{y^2}\right)=-\left(\frac{x^2+y^2}{y^3}\right)=-\frac{1}{K^2{y}^3}$

Substituting these value of $y_2$ and $y_1$ in the given expression; we get

$\frac{\left|y_2\right|}{\sqrt{{(1+{\left(y_1\right)}^2)}^3}}=\frac{|-\frac{1}{K^2{y}^3}|}{\sqrt{{(1+{(-\frac{x}{y})}^2)}^3}}=\frac{1}{K^2{y}^3}\times \frac{1}{\sqrt{{\left(\frac{x^2+y^2}{y^2}\right)}^3}}$

$=\frac{1}{K^2{y}^3}\times \frac{1}{\frac{R^3}{y^3}}=\frac{1}{K^2{R}^3}=\frac{1}{K^2}\times K^3=K$

Since $\Rightarrow { y }_{ 1 }=-\frac { x }{ y } $ substitute in given equation we get,

$\frac{y_1}{x\sqrt{1+y_1^2}}$