JEE Main2005MathematicsApplication of DerivativesActual

The normal to the curve x=a( + ), y=a( - ) at any point ' ' is such that

Options

Ait passes through the origin
Bit makes angle 2 + with the x -axis
Cit passes through ( a 2 ,- a )
Dit is at a constant distance from the origin

Correct answer

D. it is at a constant distance from the origin

Step-by-step solution

Clearly dy dx = slope of normal =- Equation of normal at ' ' is y-a( - )=- (x-a( + ) y -a ^2 +a =-x +a ^2 +a x +y =a Clearly this is an equation of straight line which is at a constant distance 'a' from origin.

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