Homework Help?

Given the curve x + xy + 2y^2 = 6\

a) find an expression for the slope of the curve at any point (x, y) on the curve

b) write an equation for the line tangent to the curve at the point (2,1)

c) find the coordinates of all other points on this curve with slope equal to the slope at (2,1)

help would be deeply appreciated

Top 1 Answers

A few days ago

Anonymous

Favorite Answer

a) Differentiating both sides w.r.t. x, we get:

1+y+xy’+4yy’=0

Now solving for y’, we find a formula for the slope:

y’=-(1+y)/(x+4y)

b) at (2,1) the slope is y’=-(1+1)/(2+4)=-1/3

Therefore, the equation of the tangent is:

(y-1)=-1/3(x-2)

c) the coordinates of such points must satisfy equation of the curve itself: x + xy + 2y^2 = 6

and

-(1+y)/(x+4y)=-1/3

y=3-x

Substituting y=3-x in the equation of the curve, we finally get:

x^2-8x+12=0

Solving for x, yields:

x=2 or x=6

Corresponding values of y are 1 and -3, respectively.

So the points in question are: (2,1) and (6,-3)

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