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For the curve x² + 2xy − 3y² = 0, what happens to the normal drawn at the point (1,1)?
- It intersects the curve again in the third quadrant.
- It intersects the curve again in the fourth quadrant.
- It does not intersect the curve again.
- It intersects the curve again in the second quadrant.
Correct answer: It intersects the curve again in the fourth quadrant.
Solution
x^2+2xy-3y^2=(x-y)(x+3y)=0, two lines. (1,1) lies on y=x; the normal there is y=-x+2. Intersecting x+3y=0 gives x=3, y=-1, i.e. the fourth quadrant.
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