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Find the equation of the tangent to the curve for the following equation at the point $(2, -3)$ $$4x^2-3xy-y^2=25$$

$$\therefore8x-3x\frac{dy}{dx}-3y-2y\frac{dy}{dx}=0 $$

$$\therefore\frac{dy}{dx} = \frac{8x-3y}{2y+3x}=\frac{16+9}{-6+6}=\frac{25}{0}$$

This is obviously undefined; however, the answer to this question is given to be $5x-2y=16$, suggesting that the gradient at the point is $2.5$. What is wrong with my working out?

StopReadingThisUsername
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    When I make a graph of your formula, I see a vertical tangent at (2,-3). Not sure where the gradient of 2.5 comes from – imranfat Apr 09 '16 at 03:32
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    I checked on Desmos, and you didn't do anything wrong. The point (2,-3) is a point of vertical tangency. Check to make sure that you copied the problem down correctly. – Jacob Apr 09 '16 at 03:32
  • Checked it myself. Thanks a lot, must be a misprint in the textbook. – StopReadingThisUsername Apr 09 '16 at 03:34

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