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Question

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


A
42
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B
22
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C
12
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D
2
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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
$$R=|\dfrac{(1+1)^{\dfrac{3}{2}}}{\dfrac{1}{2}}|$$

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

$$=4\sqrt{2}$$

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