Find an equation of the tangent line to the graph x^2 + 2y^2 + 3xy = 3 at the point (1,1)?

Answer 1

x^2 + 2y^2 + 3xy = 3
d/dx both sides
2x + 4y dy/dx + 3y + 3x dy/dx = 0
dy/dx = -(2x+3y) / (4y + 3x)
At (1,1)
dy/dx = -(2(1)+3(1)) / (4(1) + 3(1))
dy/dx = -5/7
gradient tanget to the graph = -5/7

targent pass to (1,1)
y = -5/7x + C
(1) = -5/7(1) + C
C = 12/7

Targent line : y = -5/7 x + 12/7

Answer 2

Find the tangent to the curve x² + 2y² + 3xy = 3 at (1,1).

Take the derivative implicitly.

2x + 4y(dy/dx) + 3y + 3x(dy/dx) = 0
(4y + 3x)(dy/dx) = -2x - 3y
dy/dx = (-2x - 3y)/(4y + 3x)

At the point (1,1) we have:

dy/dx = (-2 - 3)/(4 + 3) = -5/7

The slope of the tangent line is -5/7.

The equation of the tangent line to the curve at the point (1,1) is:

y - 1 = (-5/7)(x - 1) = -(5/7)x + 5/7
y = -(5/7)x + 12/7

Answer 3

x^2 + 2y^2 + 3xy = 3
d/dx both sides
2x + 4y dy/dx + 3y + 3x dy/dx = 0
dy/dx = -(2x+3y) / (4y + 3x)
At (1,1)
dy/dx = -(2(1)+3(1)) / (4(1) + 3(1))
dy/dx = -5/7
gradient tanget to the graph = -5/7

targent pass to (1,1)
y = -5/7x + C
(1) = -5/7(1) + C
C = 12/7

Targent line : y = -5/7 x + 12/7

Answer 4

Differentiate both sides:-
2x + 4y.dy/dx + 3y + 3x.(dy/dx ) = 0
[ 4y + 3x].(dy / dx) = - 2x - 3y
dy / dx = - (2x + 3y) / (4y + 3x)
dy / dx = - 5 / 7 = gradient , m at (1,1)
Equation of line thro` (1,1) is given by:-
y - 1 = (- 5 / 7).(x - 1)
7y - 7 = - 5.x + 5
5x + 7y - 12 = 0 or:-
y = (- 5/7).x + 12 / 7