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So the question is $y^2 +11xy-8x^3=-700$ find two lines tangent to the curve at the points on the curve where x=5. What is the sum of their slopes?

i got $\frac{dy}{dx}=\frac{24x^2-11y}{11x+2y}$ but how do i get the sum of their slopes at x=5?

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setting $x=5$ and solving the equation you will end up with two possible solutions for $y$:

$y=5$ and $y=60$

Now apply those two points to your differential equation to find the tangents $(5, 5)$ and $(5, 60)$ (assuming you differentiated correctly!)

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  • $\begingroup$ i got the sum as $-35/13$ is that correct? $\endgroup$ – user3467226 Oct 14 '14 at 4:12
  • $\begingroup$ it's no hard calculations, but I want first to know how did you differentiate to obtain $\frac{dy}{dx}$? $\endgroup$ – chouaib Oct 14 '14 at 4:16
  • $\begingroup$ @user3467226 how you got $24x^2-11y$ shouldn't it be $24x^2+11y$ ? $\endgroup$ – chouaib Oct 14 '14 at 4:27
  • $\begingroup$ I am weak on minus signs, but $24x^2-11y$ looks OK. $\endgroup$ – André Nicolas Oct 14 '14 at 4:43
  • $\begingroup$ @AndréNicolas, my bad !! don't know even how I managed to get the plus sign, btw I'm weaker when it comes to minus sign ;) $\endgroup$ – chouaib Oct 14 '14 at 4:50

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