i found the derivative to be dy/dx =  (2x + 2y)/(2x  6y)
then i found the slope at that point to be equal to 1, which makes the slope of the normal 1
then i found the normal line to be y = x + 2
do i set the curve equal to the line?
as in
x^2 + 2xy  3y^2 = 0
and y  x + 2 = 0
so
x^2 + 2xy  3y^2 = y  x + 2
i did this, then i tried to solve for y. i couldn't. i don't think i'm on the right track though.
so i guess, i'm not sure what to do after i get the normal line
is it (2/3,2/9) ???
3 Answers

No I don't think you were on the right track.
You can rearrange the equation into two squares as follows:
x^2 + 2xy  3y^2 = 0
x^2 + 2xy + y^2 = 4y^2
(x + y) ^ 2 = (2y) ^ 2
x + y = +/ 2y
x = y or  3y
y = x or x/3
Graphically, the 'curve' is two straight lines passing through the origin.
The point (1,1) lies on one of these lines, the line y = x. The slope of this line = 1. The slope of the normal must be 1. Therefore the normal at (1,1) has the equation
y  1 = 1 (x  1) = 1  x
y = 2  x
The normal clearly does not intersect the line y = x at any other point. It would however intersect the line y = x/3. To find this intersection point we would simply solve
y = 2  x and y = x/3
x/3 = 2  x
x = 6  3x
2x = 6
x = 3
y = 1
i.e. the normal at (1,1) also intersects the curve at (3, 1).

x² + 2xy  3y² = 0
Take the derivative and evalutate it at the point (1, 1).
2x + 2y + 2x(dy/dx)  6y(dy/dx) = 0
x + y + x(dy/dx)  3y(dy/dx) = 0
(x  3y)(dy/dx) = x  y
dy/dx = (x  y)/(x  3y) = (x + y)/(3y  x)
dy/dx = (1 + 1)/(3*1  1) = 2/2 = 1
The slope of the equation of the line normal to the curve at (1,1) is the negative reciprocal or m = 1. The equation of the normal line is:
y  1 = (x  1) = x + 1
y = x + 2
_______
Plug the value for y into the equation of the curve and solve for x.
x² + 2xy  3y² = 0
x² + 2x(x + 2)  3(x + 2)² = 0
x²  2x² + 4x  3(x²  4x + 4) = 0
x²  2x² + 4x  3x² + 12x  12 = 0
4x² + 16x  12 = 0
x²  4x + 3 = 0
(x  1)(x  3) = 0
x = 1, 3
______
Plug the value of x into the curve and solve for y. Plug in x = 1.
x² + 2xy  3y² = 0
1 + 2y  3y² = 0
3y²  2y  1 = 0
(3y + 1)(y  1) = 0
y = 1/3, 1
Only one of these values is on the normal line. The other solution is extraneous. The only solution is:
y = 1
So one point at the intersection of the curve and the normal line is (1, 1).
_______
Plug the value of x into the curve and solve for y. Plug in x = 3.
x² + 2xy  3y² = 0
9 + 6y  3y² = 0
3  2y + y² = 0
y²  2y  3 = 0
(y + 1)(y  3) = 0
y = 1, 3
Only one of these values is on the normal line. The other solution is extraneous. The only solution is:
y = 1
So a second point at the intersection of the curve and the normal line is (3, 1).
_______
The two points of intersection are:
(1, 1) and (3, 1)

The equation x² + y²  4x + 0y  9 = 0 i.e., ( x  2 )² + ( y  0 )² = 13 = (?13)² represents a circle with C(2,0) and r = ?(13). _____________________________________ Line L : 3x  2y + ok = 0 intersects the circle in 2 factors. hence, if seg CP is perpendicular to L, then ... CP < r ?  3(2)  2(0) + ok  / ?(3²+(2)²) < ?13 ?  ok + 6  < 13 ? 13 < (ok+6) < 13 ? 136 < (ok+6)6 < 136 ? 19 < ok < 7 ......................... Ans. ______________________________________... ........... satisfied to assist ! ______________________________________...