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  • Let  then the slope of normal at (4,1) on the curve is <

Let y^3 - yx + 3= 0 then the slope of normal at (4,1) on the curve is 

Option: 1

-\frac{1}{2}


Option: 2

1


Option: 3

\frac{3}{2}


Option: 4

2


Answers (1)

best_answer

Differentiating the equation of the curev w.r.t. x

\Rightarrow 3y^2 \frac{dy}{dx}- y - x\frac{dy}{dx}= 0 \Rightarrow \frac{dy}{dx}= \frac{y}{3y^2-x }

Now, \frac{\mathrm{d} y}{\mathrm{d} x} at (4,1) = \frac{1}{3 - 4} = -1= slope of the tangent  at (4,1)

Therefore, slope of normal  = 1

Posted by

Divya Prakash Singh

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