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
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.