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Question:

The normal to the curve x² + 2xy - y² = 0 at (1, 1) meets the curve again in which quadrant?

meets the curve again in the second quadrant

meets the curve again in the fourth quadrant

does not meet the curve again

meets the curve again in the third quadrant

Solution:

x² + 2xy - y² = 0
This is the equation of a pair of straight lines (x - y)(x + 3y) = 0
2x + 2y + 2xy' - 2yy' = 0 ⇒ x + y + xy' - yy' = 0
At (1, 1) the slope of the tangent is y' = -1.
The slope of the normal is 1.
Hence, the normal has the equation: x - y = 0
We need to find its intersection with x + 3y = 0
On solving we get, x = y and x + 3x = 0 ⇒ 4x = 0 ⇒ x = 0, y = 0
This is incorrect. Let's use the equation x + y = 2 instead.
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.