The normal to the curve at
- A.does not meet the curve again
- B.meets in the curve again the second quadrant
- C.meets the curve again in the third quadrant
- D.meets the curve again in the fourth quadrant
Show answer and solution
Correct option: D
Solution
Factoring x² + 2xy - 3y² = 0 gives (x - y)(x + 3y) = 0, so the curve is two lines: y = x and x + 3y = 0.
The point (1,1) lies on y = x. The normal at (1,1): implicit differentiation gives 2x + 2y + 2xy' - 6yy' = 0, so at (1,1): 2 + 2 + 2y' - 6y' = 0 → y' = 1. The normal has slope -1, giving equation y - 1 = -(x - 1), i.e., y = -x + 2.
Find intersection with the other line x + 3y = 0 (i.e., y = -x/3):
-x/3 = -x + 2 → -x = -3x + 6 → 2x = 6 → x = 3, y = -1.
The point (3, -1) is in the fourth quadrant.
The point (1,1) lies on y = x. The normal at (1,1): implicit differentiation gives 2x + 2y + 2xy' - 6yy' = 0, so at (1,1): 2 + 2 + 2y' - 6y' = 0 → y' = 1. The normal has slope -1, giving equation y - 1 = -(x - 1), i.e., y = -x + 2.
Find intersection with the other line x + 3y = 0 (i.e., y = -x/3):
-x/3 = -x + 2 → -x = -3x + 6 → 2x = 6 → x = 3, y = -1.
The point (3, -1) is in the fourth quadrant.
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