Equation of the tangent

[Mel Mitch]

Hello again....

Find the equations of the curve y^2 + 3xy +4x^2= 14 at the points where x=1.
Anwers y+x+4=0, y+2x =4

i use implicit methods

2y(dy/dx) + 3x(dy/dx)+8x=0
dy/dx= -8x/(2y+3x)

or do i substitude x=1 in the equation first equation y^2+3y+4=14,factorize it Y=2 Y=-5
????? help

 

[arthur ohlsten]

y^2+3xy+4x^2=14
find the y value at x=1

y^2+3y+4=14
y^2+3y-10=0
[y+5][y-2]=0
y=-5 or y=2
at x=1 y=-5 or at x=1 y=2

find the tangent line at x=1 y=-5 AND at x=1 y=2

y=mx+b
find the slope dy/dx at x=1 y=-5 this is m term
2 y dy/dx +3[x dy/dx + y ] +8x=0
dy/dx[2y +3x] = -8x-3y
dy/dx = {-8+15]/[-10+3]
dy/dx = 7/[-7]
dy/dx =-1 or m=-1

y=-x+b but x=1 y=-5
-5=-1 + b
b=-4
y=-x-4 tangent line answer one

you can slove for x=1 y=2

Arthur

 

[BigGlenntheHeavy]

You have an "oblique" ellipse with two linear equations where x = 1, namely y=-2x+4 and y = -x-4.
Both linear equations are tangent to the graph of the ellipse at x = 1. Equation of ellipse is y^2+3xy+4x^2=14.

Note: Observe graph.[attachment=0:2t4zai6m]http://www.jpg[/attachment:2t4zai6m]

 

[Mel Mitch]

yes i can arthur....thank u.....and thanks Big G