JEE Main 2015 Mathematics question

MathematicsApplication of DerivativesJEE Main 2015

The distance from the origin, of the normal to the curve, , at , is :

Answer: (4)

Solution

Differentiate with respect to : and . Hence , so at the tangent slope is 1 and the normal slope is . The point of contact is $ = 2ยทMissing argument for \frac\frac1{Missing argument for \sqrt\sqrt2} +

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Question ID VK-115305