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Question

The equation of the normal to the curve y4=ax3 at a,a is


A

x+2y=3a

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B

3x-4y+a=0

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C

4x+3y=7a

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D

4x-3y=0

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Solution

The correct option is C

4x+3y=7a


Explanation for the correct option

Step 1: Solve for the slope of normal

Given curve is, y4=ax3 and the point is a,a

Differentiating it implicitly,

4y3dydx=3ax2⇒dydx=3ax24y3

Where dydx is the slope of the tangent

At a,a, x=a and y=a. Substituting this in above equation,

⇒dydx=3a·a24a3⇒dydx=34

We know that,

Slope of normal = (Slope of tangent)-1

=-1dydx=-134=-43

Step 2: Solve for the equation of normal

Equation of a line in point-slope form is

y-y1=mx-x1
where x1,y1 is the point the line passes through and m is the slope.

For the normal of the given curve here, x1,y1=a,a and m=-43.

Thus the equation of the normal is,

y-a=-43x-a⇒3y-3a=4a-4x⇒4x+3y=7a

Therefore, the equation of the normal to the given curve is 4x+3y=7a.

Hence, option(C) is correct.


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