Implicit functions

[Mel Mitch]

Given that y^2 - 5xy + 8x^2 = 2, prove that dy/dx= (5y - 16x)/ (2y - 5x).

The distinct points P and Q on the curve y^2 -5xy +8x^2 = 2 each have x- coordinate 1. The normals to the curve at P and meet at the point N. Calcutate the coordinates of N.

My steps
2y (dy/dx) - 5x(dy/dx) + 16x = 0

dy/dx (2y-5x)= -16x

dy/dx= -16/(2y-5x)

How do i get the 5y infront of the -16x...what i'm i doing wrong.

 

[arthur ohlsten]

y^2-5xy+8x^2=2
find dy/dx

2y dy/dx -5 [x dy/dx +y dx/dx]+16x dx/dx=0 dx/dx =1 [only done to show your error]
2y dy/dx -5[x dy/dx +y]+16x=0
dy/dx [ 2y-5x]=[5y-16x]
dy/dx = [5y-16x] / [2y-5x] proof
=============================================
find points P and Q
P and Q are at x=1 y=?
let x=1
y^2-5y+8=2
y^2-5y+6=0
[y-3][y-2]=0
y=3 or y=2

point P x=1 y=3
point Q x=1 y=2
====================================================================
find slope of curve at P and Q
slope at P , dy/dx = [15-16]/[6-5]
dy/dx = -1
then slope of curve normal at P is m=1

curve is y=mx+b
y=x+b but at x=1 y=3
3=1+b
b=2
y=x+2 answer one

you solve for normal at Q,x=1 y=2


please check algebra
Arthur

 

[BigGlenntheHeavy]

\(\displaystyle f(x,y) \ = \ y^{2}-5xy+8x^{2}-2 \ = \ 0\)

\(\displaystyle f(1,y) \ = \ y^{2}-5y+6 \ = \ (y-3)(y-2) \ = \ 0\)

\(\displaystyle f ' (x,y) \ = \ 2yy'-5(y+xy')+16x \ = \ 0\)

\(\displaystyle m \ = \ y' \ = \ \frac{5y-16x}{2y-5x}\)

\(\displaystyle Thus, \ when \ x \ = \ 1 \ and \ y \ = \ 2, \ m \ = \ 6, \ then \ y \ = \ 6x-4 \ (red \ line) \ and\)
\(\displaystyle \ NORM \ y \ = \ (-x+13)/6 \ (blue \ line), \ see \ graph.\)

\(\displaystyle When \ x \ = \ 1 \ and \ y \ = \ 3, \ then \ m \ = \ -1 \ and \ y \ = \ -x+4 \ (black \ line) \ and\)
\(\displaystyle NORM \ y \ = \ x+2 \ (green \ line).\)

\(\displaystyle Both \ NORMS \ intersect \ at \ N \ = \ (\frac{1}{7},\frac{15}{7}).\)

Note: Both NORMALS are normal to their respective lines at the point of tangency of those lines.

[attachment=0:2w5apdx8]bbb.jpg[/attachment:2w5apdx8]

 

[Deleted member 4993]

Given that y^2 - 5xy + 8x^2 = 2, prove that dy/dx= (5y - 16x)/ (2y - 5x).

The distinct points P and Q on the curve y^2 -5xy +8x^2 = 2 each have x- coordinate 1. The normals to the curve at P and meet at the point N. Calcutate the coordinates of N.

My steps
2y (dy/dx) - 5x(dy/dx) - 5y + 16x = 0

(dy/dx)(2y -5x) = 5y - 16x

dy/dx = (5y - 16x)/(2y - 5x)

How is it that you are in calcIII - and cannot arrive at this "given" answer!!!

dy/dx (2y-5x)= -16x

dy/dx= -16/(2y-5x)

How do i get the 5y infront of the -16x...what i'm i doing wrong.

 

[Mel Mitch]

Thanks so much Big G and arthur...i completed my other implicit questions correctly.....thanks again