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

$\frac{{y''}}{{\sqrt {1 + y'} }}$

$\frac{{2y''}}{{\sqrt {1 + {{y'}^2}} }}$

$\frac{{y''}}{{2\sqrt {{{\left( {1 + {{y'}^2}} \right)}^3}} }}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

{x^2} + {y^2} = {a^2}

and

k = 1/a

, then

k

is equal to

View answer and explanation

Correct option: B

\frac{{\left| {y''} \right|}}{{\sqrt {{{\left( {1 + {{y'}^2}} \right)}^3}} }}

The correct answer is B (

\frac{{\left| {y''} \right|}}{{\sqrt {{{\left( {1 + {{y'}^2}} \right)}^3}} }}

) because it represents the radius of curvature

k = \frac{1}{a}

derived from the given equation and its derivatives.

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