Calculus graph?

Question

The graph of (x^2) + 4y^2 - 4x -12y + 4 = 0 has two points of horizontal tangency, (x1,y1) and (x2,y2) Find the value of (x1+y1+x2+y2)(x1+x2).

(x1 and the like should be read as x sub 1.)

I could really use an explanation on how to do this, not just an answer.

Answer 1

Find horizontal tangents by setting y ' = 0
Then use those x-values in original equation to get y values
Implicit Differentiation to get y '

2x + 8 y y' - 4 - 12 y' = 0 ===> solve for y'
8 y y' - 12 y' = 4 - 2x
(8y-12) y' = 4 - 2x
y ' = (4-2x) / (8y-12)

set y' = 0 ===> means numerator is 0
4-2x=0 ==> x=2 (places where y'=0)

Plug in x=2 into original to get the corresponding points.
(x^2) + 4y^2 - 4x -12y + 4 = 0
(4) + 4y^2 - 8 -12y + 4 = 0
4y^2 - 12y = 0
4y ( y-3) = 0 ====> y=0 or y=3

The points of horizontal tangency are
(2,0) and (2,3)
hence x1=2 and x2=2 and y1=0 and y2=3
plugging into (x1+y1+x2+y2)(x1+x2)
(2+0+2+3)(2+2) = 7*4 = 28

=]
good luck!


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Answer 2

(x^2) + 4y^2 - 4x -12y + 4 = 0
x^2-4x +4(y^2-3y) +4 = 0
Now complete the square in x an y:
x^2-4x +4 +4(y^2-3y +9/4) = 4+9 =13
(x-2)^2 +4(y-3/2)^2 = 13
This is the equation of a circle with center at (2, 1.5) and radius = sqrt(13)

The horizontal tangents will occur when x = 2.
So 4 + 4y^2 -8 -12y +4 = 0
4y^2 -12y= 0
y^2 -3y = 0
y = 0,3
So x1=x2 = 2, y1 =0 and y2 = 3
(x1+x2+y1+y2)(x1+x2+ = (2+2+0+3)/2+2)= 7*4=28