Consider the equation x^2-2xy+4y^2=64. Find the equation of the tangent lines to the curve at the point x=2??

Answer 1

Take the derivative implicitly.
2x - 2y - 2x y' + 8y y' = 0
Move everything that does not have y' to the other side
-2x y' + 8y y' = 2y - 2x
Factor y' on the left
y'(-2x+8y) = 2y - 2x
Divide both sides by (-2x+8y)
y' = (2y-2x) / (-2x+8y)
Divide everything by 2
y' = (y-x) / (-x+4y) = (y-x) / (4y-x)
You are given that x = 2, now you must find the corresponding y value(s).
If x = 2, then
(2)^2 - 2(2)y + 4y^2 = 64
4 - 4y + 4y^2 = 64
Divide everything by 4
1 - y + y^2 = 16
Subtract 16 from each side and rearrange.
y^2 - y - 15 = 0
Solve this to find your y values.
You will have to points (x,y).
Subsitute each pair into the equation for y'. This will give you the slope, then write the equation of the line.

Answer 2

take the spinoff with d/dx... remembering that d/dx ( f(x) ) = f ' (x) dx / dx = f ' (x) and d /dx ( f(y) ) = f ' (y) dy/dx occasion : d / dx (x^2) = 2x yet d / dx (y^2) = 2y dy/dx first and third words are potential regulations, however the middle is a product rule (u'v + uv') 4x + 5(y + x*dy/dx) + 10y*dy/dx = 0 4x + 5y + 5x*dy/dx + 10y*dy/dx = 0 4x + 5y + (5x + 10y)*dy/dx = 0 plug interior the factor (2,a million) it fairly is (x,y) 4(2) + 5(a million) + (5(2) + 10(a million))dy/dx = 0 8 + 5 + 20*dy/dx = 0 dy/dx = -13/20 so we've the slope and the factor, use the factor slope style from algebra (y-y1) = m(x-x1) y-a million = -13/20(x-2) y = -13/20 (x-2) + a million... i might go away it purely how that is

Answer 3

already answered this one! it doesn't get easier when you re-ask it! those square roots aren't going to get "lost" once you ask it the tenth time or anything like that..........