Application of DerivativeEasy
Question
The normal to the curve, x2 + 2xy - 3y2 = 0, at (1, 1)
Options
A.meets the curve again in the third quadrant
B.meets the curve again in the fourth quadrant
C.does not meet the curve again
D.meets the curve again in the second quadrant
Solution
x2 + 3xy - xy - 3y2 = 0
x(x + 3y) - y(x + 3y) = 0
(x - y) (x + 3y) = 0
Normal at (1, 1) will be x + y = 2
Now, x + y = 2
x + 3y = 0
2y = - 2, y = -1
& x = 3
∴ (3, - 1)
which is in fourth quadrant.
x(x + 3y) - y(x + 3y) = 0
(x - y) (x + 3y) = 0
Normal at (1, 1) will be x + y = 2
Now, x + y = 2
x + 3y = 0
2y = - 2, y = -1
& x = 3
∴ (3, - 1)
which is in fourth quadrant.
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