at a particular point of the curve 5x^2 -12x+1 the equation of the normal is x+18y+c=0. find the value of the constant c.?

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answers:
Φ² = Φ+1 say: The normal is x+18y+c=0.
The slope of the normal to y=5x^2-12x+1 is 1/(12-10x) which equals -1/18.
Here at the normal point this is (x₀=(18+12)/10, y₀=5((18+12)/10)^2 - 12(18+12)/10 + 1) ~ for you to simplify.
From this c = -x₀ - 18y₀ ~ for you to evaluate.
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alex say: y=5x^2 -12x+1
y' = 10x-12 = 18 --->x = 3 , y = 10
3+18(10)+c=0 --->c = ...
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mifop say: txjzzxpf
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Captain Matticus, LandPiratesInc say: The normal will be -dx/dy

y = 5x^2 - 12x + 1
dy = 10x * dx - 12 * dx
dy = (10x - 12) * dx
1 / (10x - 12) = dx/dy
1 / (12 - 10x) = -dx/dy

x + 18y + c = 0
18y = -x - c
y = (-1/18) * x - (1/18) * c

The slope of the normal is -1/18

1 / (12 - 10x) = -1 / 18
18 = -(12 - 10x)
18 = 10x - 12
30 = 10x
3 = x

5 * 3^2 - 12 * 3 + 1 =>
5 * 9 - 36 + 1 =>
45 - 36 + 1 =>
10

x = 3 , y = 10

3 + 18 * 10 + c = 0
3 + 180 + c = 0
183 + c = 0
c = -183
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