# Equation of line normal to curve

• Jan 24th 2013, 08:06 AM
togo
Equation of line normal to curve
Hi. Equation for curve:
y = 6x - 2x^2 at (2,4)

My work:
= 6 - 4x
= 6 - 4(2)
= 6 - 8 = -2 = M (slope)

reciprocal: 1/-2
y - y1 = m (x - x1)
y - 4 = -2(x-2)
y - 4 = -2x + 4
y = -2x

but this is wrong? thanks.
• Jan 24th 2013, 09:00 AM
AZach
Re: Equation of line normal to curve
The normal line is the line perpindicular to the curve, and that requires the negative reciprocal I think. So the slope of the normal line should be just 1/2. You can type the question into WolframAlpha to confirm.
• Jan 24th 2013, 09:32 AM
togo
Re: Equation of line normal to curve
got it. thanks
• Jan 24th 2013, 03:26 PM
Prove It
Re: Equation of line normal to curve
Quote:

Originally Posted by togo
Hi. Equation for curve:
y = 6x - 2x^2 at (2,4)

My work:
= 6 - 4x
= 6 - 4(2)
= 6 - 8 = -2 = M (slope)

reciprocal: 1/-2
y - y1 = m (x - x1)
y - 4 = -2(x-2)
y - 4 = -2x + 4
y = -2x

but this is wrong? thanks.

It's already been established that the gradient of your normal is 1/2, not -1/2. But I'm wondering why you substituted -2 as m?