Calculus problem involving slope?

Question

given the curve x+xy+2y^2=6

i found the derivative

y'=-1-y/x+4y

and i found the equation tangent to curve

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

but the question also asks me

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

Answer 1

At (2,1), slope = 1-1/2+4*1= 9/2 plugging in to slope formula

Thus 9/2= 1-y/x+4y

7/2=-y/x+4y

it can be solved for either variable

Answer 2

It's asking you to find all values of (x, y) where

the slope is equal to the slope at (2, 1), which I

assume is the -1/3 in your linear equation.

(I didn't check your work.)

So, you have two equations and two unknows.

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

and

x + xy + 2y^2 = 6

Solve the two equations together to get your values of x and y.

Answer 3

so far so good.  slope at (2,1) = -1/3, so solve

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

x + 4y = 3 + 3y

y = -x + 3

and plugging that back into the original curve,

x + x(-x+3) + 2(-x+3)² = 6

x - x² + 3x + 2(x² - 6x + 9) = 6

x - x² + 3x + 2x² - 12x + 18 = 6

x² - 8x + 12 = 0

(x - 2)(x - 6) = 0

if x = 2, y = 3-2 = 1, so (2,1) our original point.

if x = 6, y = 3-6 = -3, so (6,-3) is the other point.