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junior college 1 | H2 Maths
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hi can you help me with part i thanks
Date Posted: 3 years ago
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One possible equation of the hyperbola is (x - h)² / a² - (y - k)² / b² = 1, whereby :
→The centre has coordinates (h,k)
→ a is the distance from the centre to the vertex
→ b is the distance from the centre to the co-vertex
Since the origin (0,0) is at the centre of the base, and the narrowest part of the tower (which the centre has the same y-coordinate as) is at 80m (80units) above the centre,
We can deduce the centre has coordinates (0,80). Therefore h = 0 and k = 80 . We also know the vertices each have y-coordinate of 80.
Since the circular base has a radius of 50m, we can deduce the points (-50,0) and (50,0) lie on the hyperbola.
Sub (50,0) and k = 80 into the equation,
50² / a² - (0 - 80)² / b² = 1
2500/a² - 6400/b² = 1
2500 - 6400a²/b² = a²
2500 = a² + 6400a²/b²
2500 = a²(1 + 6400/b²)
a² = 2500/(1 + 6400/b²)①
Since the circular top has a radius of 37.5m, and the top is 125m above the base, we can deduce the points (-37.5,125) and (37.5,125) lie on the hyperbola.
Sub (37.5,125) and k = 80 into the equation,
(37.5)² / a² - (125 - 80)² / b² = 1
1406.25/a² - 2025/b² = 1 ②
Solving for ① and ② (either using GC or manual), b² = 3600, a² = 900
→The centre has coordinates (h,k)
→ a is the distance from the centre to the vertex
→ b is the distance from the centre to the co-vertex
Since the origin (0,0) is at the centre of the base, and the narrowest part of the tower (which the centre has the same y-coordinate as) is at 80m (80units) above the centre,
We can deduce the centre has coordinates (0,80). Therefore h = 0 and k = 80 . We also know the vertices each have y-coordinate of 80.
Since the circular base has a radius of 50m, we can deduce the points (-50,0) and (50,0) lie on the hyperbola.
Sub (50,0) and k = 80 into the equation,
50² / a² - (0 - 80)² / b² = 1
2500/a² - 6400/b² = 1
2500 - 6400a²/b² = a²
2500 = a² + 6400a²/b²
2500 = a²(1 + 6400/b²)
a² = 2500/(1 + 6400/b²)①
Since the circular top has a radius of 37.5m, and the top is 125m above the base, we can deduce the points (-37.5,125) and (37.5,125) lie on the hyperbola.
Sub (37.5,125) and k = 80 into the equation,
(37.5)² / a² - (125 - 80)² / b² = 1
1406.25/a² - 2025/b² = 1 ②
Solving for ① and ② (either using GC or manual), b² = 3600, a² = 900
Solving manually :
2500/a² - 6400/b² = 1 ①
1406.25/a² - 2025/b² = 1 ②
① = ②
2500/a² - 6400/b² = 1406.25/a² - 2025/b²
2500/a² - 1406.25/a² = 6400/b² - 2025/b²
1093.75/a² = 4375/b²
Cross multiply,
1093.75b² = 4375a²
b² = 4375a² / 1093.75
b² = 4a²
Sub this into ①,
2500/a² - 6400/4a² = 1
2500/a² - 1600/a² = 1
900/a² = 1
a² = 900
So b² = 4(900) = 3600
2500/a² - 6400/b² = 1 ①
1406.25/a² - 2025/b² = 1 ②
① = ②
2500/a² - 6400/b² = 1406.25/a² - 2025/b²
2500/a² - 1406.25/a² = 6400/b² - 2025/b²
1093.75/a² = 4375/b²
Cross multiply,
1093.75b² = 4375a²
b² = 4375a² / 1093.75
b² = 4a²
Sub this into ①,
2500/a² - 6400/4a² = 1
2500/a² - 1600/a² = 1
900/a² = 1
a² = 900
So b² = 4(900) = 3600
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