Find Equations for Differentials of 9y^2=x^3+3x^2 at (1,2/3), (-2,2/3), (-3,0)

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The discussion focuses on finding the equations for the differentials of the curve defined by 9y^2 = x^3 + 3x^2 at specific points. The derivative is correctly identified as y' = (x^2 + 2x) / (6y). At the point (1, 2/3), the slope is 3/4, leading to the equation 9x - 12y - 1 = 0. For the point (-2, 2/3), the slope is 0, resulting in the equation y = 2/3. The slope at (-3, 0) does not exist, indicating a vertical line at x = -3, which corrects the initial interpretation.
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If anyone could check this answer for me, it would be greatly appreciated.

Find the equations relating the differentials on the curve 9y^2= x^3 +3x^2 at the points (1,\frac{2}{3}), (-2,\frac{2}{3}), (-3,0)

Here's what I got:

y'= \frac {x^2+2x}{6y}

m @ (1,\frac{2}{3})= \frac {3}{4}
equation: 9x-12y-1=0

m @ (-2,\frac{2}{3})=0
equation: y= 2/3

m @ (-3,0)
does not exist

Thanks in advance.
 
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erik05 said:
If anyone could check this answer for me, it would be greatly appreciated.

Find the equations relating the differentials on the curve 9y^2= x^3 +3x^2 at the points (1,\frac{2}{3}), (-2,\frac{2}{3}), (-3,0)

Here's what I got:

y&#039;= \frac {x^2+2x}{6y} <------ CORRECT

m @ (1,\frac{2}{3})= \frac {3}{4}
equation: 9x-12y-1=0 <----- CORRECT

m @ (-2,\frac{2}{3})=0
equation: y= 2/3 <----- CORRECT

m @ (-3,0)
does not exist <----- Should be the Vertical Line {x = -3}

Thanks in advance.
Everything appears correct except the last item.

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Thanks for the help.
 
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