Question

The radius of curvature of the curve $y^2=4x$ at the point $(1,2)$ is

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $4\sqrt{2}$

Radius of curvature

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

Now

$y^{\prime }{|}_{(1,2)}$

$=\frac{2x}{y}{|}_{1,2}$

$=1$

And

$y^{\prime \prime }{|}_{1,2}$

$=\frac{2y-2x{y}^{\prime }}{y^2}{|}_{1,2}$

$=\frac{2y-2x\frac{2x}{y}}{y^2}{|}_{1,2}$

$=\frac{2y^2-4x^2}{y^3}{|}_{1,2}$

$=\frac{8-4}{8}$

$=\frac{1}{2}$.

Therefore

$R=|\frac{(1+1{)}^{\frac{3}{2}}}{\frac{1}{2}}|$

$=\frac{2\sqrt{2}}{\frac{1}{2}}$

$=4\sqrt{2}$