Question

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

A
42
B
22
C
12
D
2

Solution

## The correct option is A $$4\sqrt{2}$$Radius of curvature $$=|\dfrac{(1+y')^{\frac{3}{2}}}{y''}|$$Now $$y'|_{(1,2)}$$$$=\dfrac{2x}{y}|_{1,2}$$$$=1$$And $$y''|_{1,2}$$$$=\dfrac{2y-2xy'}{y^{2}}|_{1,2}$$$$=\dfrac{2y-2x\dfrac{2x}{y}}{y^{2}}|_{1,2}$$$$=\dfrac{2y^{2}-4x^{2}}{y^{3}}|_{1,2}$$$$=\dfrac{8-4}{8}$$$$=\dfrac{1}{2}$$.Therefore$$R=|\dfrac{(1+1)^{\dfrac{3}{2}}}{\dfrac{1}{2}}|$$$$=\dfrac{2\sqrt{2}}{\dfrac{1}{2}}$$$$=4\sqrt{2}$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More