The normal to the curve, x2 + 2xy – 3y2 = 0, at [1, 1]

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does not meet the curve again

meets the curve again in second quadrant

meets the curve again in third quadrant

meets the curve again in the fourth quadrant

answer is B.

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Detailed Solution

x2 + 2xy – 3y2 = 0
(x – y)(x + 3y) = 0
2x + 2y + 2xy’ – 6yy’ = 0

At (1, 1)
the slope of tangent is y = 1
The slope of the normal is –1
Hence,
The normal has the equation: x + y = 2
We need to find its intersection with x + 3y = 0
On solving we get,
2 – y + 3y = 0
⇒ 2y = –2
y = –1, x = 3
Hence, it meets it in the fourth quadrant.

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