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Implicit differentiation help!!?
[x^(2)/4] +[ y^(2)/9] =1
find the lines that are tangent and normal to each curve at the given point: (1,(3/2)*sqrt(3))
Please show me how to do this . I'm so frustrated. I know that I need to differentiate and plug in numbers to a line but I can't gts the right answers.
1 Answer
- ParadigMLv 49 years agoFavorite Answer
Well first you need to differentiate:
(1/2)x^2 +(2/9)y^2 (dy/dx)=0
Then simplify to get (dy/dx) on one side. Should look like this when you're done:
(dy/dx)= (-9x)/(4y)
That is the equation for the slope at any given point on the ellipse.
So plug in the values x=1 and y= (3/2)sqrt(3) and that gives you the slope at that point. So:
(dy/dx)= -9/(6sqrt(3))
Then using point-slope form (y-y1=m(x-x1)) you can get the equation of the tangent line. So:
y-(3/2)sqrt(3)=(-9/6sqrt(3))(x-1)
You can leave it like that, or you can change it to y=mx+b format. Both are considered equations for lines.
Then to find the normal line you take your slope (-9/6sqrt(3))x and take the reciprocal of it and multiply it by -1. The rest of the equation stays the same.
So your slope is now:
((6sqrt(3))/9)x
