Hello, can someone help me solve this problem? I tried working it around but I can't arrive at the correct answer. Thank you!
Problem: Find the equation of the curve for which y''=12/x3 if it passes through (1,0) and is tangent to the line 6x+y=6 at that point.
Answer: xy+6x=6
Source: Elements of Calculus and Analytic Geometry by Reyes and Chua
$6x + y = 6$
$6 + y' = 0$
$y' = -6$
$y''=12x^{-3}$
$y' = -6x^{-2} + C_1$ → Eq (1)
$y = 6x^{-1} + C_1x + C_2$ → Eq (2)
At the point of tangency (1, 0), y' = -6
$-6 = -6 + C_1$
$C_1 = 0$
From Eq (2)
$y = 6x^{-1} + C_2$
$0 = 6 + C_2$
$C_2 = -6$
Thus,
$y = \dfrac{6}{x} - 6$ → Eq (2)
$xy = 6 - 6x$
$xy + 6x = 6$ answer
Got it now, thanks! ;)
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