Question 1189323
Find the tangent line to the parabola x^2 = 6y + 10 through (7,5)
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y = (x^2 - 10)/6
y' = x/3 (the slope of lines tangent)

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3y = 10x - 55
and
3y = 4x - 13
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More later.
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Thru (7,5) ---> y-5 = (x/3)(x - 7)
y-5 = (x^2 - 7x)/3
3y - 15 = x^2 - 7x
y = (x^2 - 7x + 15)/3
y = (x^2 - 10)/6
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(x^2 - 7x + 15)/3 = (x^2 - 10)/6
2x^2 - 14x + 30 = x^2 - 10
x^2 - 14x + 40 = 0
(x-4)*(x-10) = 0
x = 4, x = 10 --- (ordinates of the tangent points)
x = 4 ---> (4,1)
x = 10 --> (10,15)
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For (4,1) and (7,5):
m = 4/3
y-5 = (4/3)*(x-7)
3y - 15 = 4x - 28
3y = 4x - 13 --- One line
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For (10,15) and (7,5):
m = 10/3
y-5 = (10/3)*(x-7) 
3y - 15 = 10x - 70
3y = 10x - 55 --- 2nd line