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

First, use implicit differentiation with respect to x. You'll get:
2x - 2y - 2x(dy/dx) + 8y(dy/dx) = 0
Now move the components of the equation about and you'll get:
dy/dx = (x-y)/(x-4y) This will give the slope, but you'll also need the value of y when x=2.
By substituting x=2 in the equation x^2-2xy+4y^2=64, you'll get:
4-4y+4y^2=64
This is the same as 4y^2-4y-60=0
y^2 - y 15 = 0
Using the quadratic formula, you'll find that y=4.405 and y =-3.405
When x=2 and
a) y = 4.405,
dy/dx = 0.154
Equation of tangent: y = 0.154x + 4.097
b) y = -3.405
dy/dx = 0.346
Equation of tangent: y = 0.346x-4.097

Answer 2

x^2-2xy+4y^2=64.
2x -2xy' -2y +8yy' = 0
-2xy' + 8yy' = 2y - 2x
-xy' +4yy' = y-x
y'(-x + 4y) = y-x
y' = (y-x)/(4y-x)
When x = 2,
y' = (y-2)/(4y-2)
So y'= 0 when y = 2 or y= 1/2
So slopes = 2/2 = 1 or (1/2)/2=1/4
So y= x+b1 or y= x/4 + b2
To find b1 and b2, put x =2 in original equation and solve for y, getting y= .5+/- sqrt(61)
So .5 +/- sqrt(61) = (1)(2) + b1 --> b1= .5 +/- sqrt(61) -2
So y= x + (.5 +/- sqrt(61)), and
y = x/4 + (.5 +/- sqrt(61)) are the required tangent equations.

Answer 3

So let us y(2) first. 2*2 –2*2*y +4yy =64 or yy –y –15 =0, hence there are two points: x1=2, y1=-r+.5; x2=2, y2=r+0.5, where r=sqrt(15.25);
Now let us differentiate: 2x –2y –2xy’ +8yy’ =0 or y’ = (x-y) / (x-4y);
and k= y’(x=2) = (2 +r –.5) / (2 +4r –2) = 0.375/r +0.25;
tangent line: y=k*x+b, where b to be found;
-r+0.5 = (0.375/r +0.25)*2 +b, hence b=-r -0.75/r, so y = (0.375/r +0.25)x -r-0.75/r;
line #1: y = (0.375/r +0.25)x -r-0.75/r, where r=-sqrt(15.25);
line #2: y = (0.375/r +0.25)x -r-0.75/r, where r=+sqrt(15.25);