For each curve, find the equation of the tangent at the given point. a) x^2 + y^2 = 13 at (2, -3) I got the slope, which is - x/y but I have no clue how to put this into an equation.
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Use point-slope form: $\displaystyle y-y_1 = m(x-x_1) $
I have tried that. I get y=2/3 (x) + 5/3 (correct me if I'm wrong), but the answer in the back of my textbook is 2x - 3y - 13 = 0
$\displaystyle y+3 = \frac{2}{3}(x-2) $ $\displaystyle y = \frac{2}{3}x - \frac{13}{3} $ So... $\displaystyle 3y = 2x - 13 $ Or $\displaystyle 3y-2x + 13 = 0 $ Then $\displaystyle 2x-3y-13 = 0 $
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