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Question

The point on the curve 3y=6x-5x3, the normal at which passes through the origin is,


A

1,13

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B

13,1

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C

2,-283

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D

None of these.

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Solution

The correct option is A

1,13


The explanation for the correct option.

Step-1 : Slope of the normal of the curve:

Let P(a,b) be the required point.

Given the equation of curve is 3y=6x-5x3

Differentiate with respect to x.

3dydx=6-15x2dydx=2-5x2

The product of the slope of two perpendicular lines is equal to -1.

Thus, the Slope of the tangent at P(a,b)=2-5a2

Slope of normal =15a2-2

Step-2: Equation of the normal:

Equation of normal at P is computed as,

y-b=15a2-2(x-a)

Since it passes through the origin. So substitute (x,y)=(0,0).

b=a5a2-2

When we substitute all the given options in the above equation. Then, the only option Asatisfies the above equation.

Thus, the required point is 1,13.

Hence, option A is correct.

The explanation for the incorrect option.

Since the option B,C,andD does not satisfy the equation b=a5a2-2.

Hence, the option B,C,andD are incorrect.

Hence, option A is correct.


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