# Finding equation of the tangent to the curve

### Finding equation of the tangent to the curve

I need to find the equation of the tangent to the curve xy2 + 3x -3y=6. at the point (2,1)

I know i need to differentiate implicitly but its getting confusing differentiating the xy2?
After differentiation then id set the function dy/dx to equal 0?

Guest

### Re: Finding equation of the tangent to the curve

sorry i made a typo its xy2 + 3x -2y=6. at the point (2,1)
Guest

### Re: Finding equation of the tangent to the curve

Why don't you solve the equation for y. Then you get $y=\frac{sqrt{-3x^2+6x+1}+1}{x}$ and $y=\frac{-sqrt{-3x^2+6x+1}+1}{x}$. The point (2,1) lies on the first function. So you have to find the tangent to $f(x)=\frac{sqrt{-3x^2+6x+1}+1}{x}$. The last is easy. Calculate $f'(x)$ at point (2,1) and you get the the "a" of ax+b=0 - the equation of the tangent. (I think you know these and you know haw to proceed to find b)

MM

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