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junior college 2 | H3 Maths
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Date Posted: 6 years ago
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3y² - 2qy + 2y + q² - 2q - 3 = 0
3y² + y(2 - 2q) + q² - 2q - 3 = 0
Discriminant = b² - 4ac
= (2-2q)² - 4(3)(q² - 2q - 3)
= 4 - 8q + 4q² - 12q² + 24q + 36
= 16 q + 40 - 8q²
= 8 (2q + 5 - q²)
If q² < 2q + 5,
Then 2q + 5 > q²
2q + 5 - q² > 0
So 8(2q + 5 - q²) also > 0
So 16q + 40 - 8q² > 0
Thus, discriminant > 0
So the line will intersect the ellipse at two distinct(different) points.
3y² + y(2 - 2q) + q² - 2q - 3 = 0
Discriminant = b² - 4ac
= (2-2q)² - 4(3)(q² - 2q - 3)
= 4 - 8q + 4q² - 12q² + 24q + 36
= 16 q + 40 - 8q²
= 8 (2q + 5 - q²)
If q² < 2q + 5,
Then 2q + 5 > q²
2q + 5 - q² > 0
So 8(2q + 5 - q²) also > 0
So 16q + 40 - 8q² > 0
Thus, discriminant > 0
So the line will intersect the ellipse at two distinct(different) points.
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