SOLUTION: find the equation of normal to the curve 3ay2= x2(x+a) at (2a.2a

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Question 1076007: find the equation of normal to the curve 3ay2= x2(x+a) at (2a.2a
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
3ay%5E2=x%5E2%28x%2Ba%29
Implicitly differentiate both sides,
6aydy=x%5E2dx%2B%28x%2Ba%29%282xdx%29
6aydy=%28x%5E2%2B2x%28x%2Ba%29%29dx
dy%2Fdx=%28x%5E2%2B2x%28x%2Ba%29%29%2F%286ay%29
When x=2a,
dy%2Fdx=%28%282a%29%5E2%2B2%282a%29%282a%2Ba%29%29%2F%286a%282a%29%29
dy%2Fdx=%284a%5E2%2B12a%5E2%29%2F%2812a%5E2%29
dy%2Fdx=%2816a%5E2%29%2F%2812a%5E2%29
dy%2Fdx=4%2F3
The normal is perpendicular to the tangent and the slopes are negative reciprocals,
m%5BN%5D%2A%284%2F3%29=-1
m%5BN%5D=-3%2F4
So then using the point-slope form,
y-2a=-%283%2F4%29%28x-2a%29
4%28y-2a%29=-3%28x-2a%29
4y-8a=-3x%2B6a
highlight%283x%2B4y=14a%29