the line that is normal to the curve x^2 + 2xy – 3y^2 = 0 at (1,1) intersects the curve at what other point?

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

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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

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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).

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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).

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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 ? -13-6 < (ok+6)-6 < 13-6 ? -19 < ok < 7 ......................... Ans. ______________________________________... ........... satisfied to assist ! ______________________________________...