Finding equation of the tangent to the curve

by **Guest** » Thu Jul 05, 2012 1:15 pm

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?

by **Guest** » Thu Jul 05, 2012 1:17 pm

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

by MM » Fri Jul 06, 2012 2:17 pm

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)

Finding equation of the tangent to the curve

Finding equation of the tangent to the curve

Re: Finding equation of the tangent to the curve

Re: Finding equation of the tangent to the curve

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