the focal point of an ellipse are (12, 0) and (-12, 0), and the point (12,7 ) is on the ellipse.Find the point where this curve intersects the coordinate axes
Please show steps, and explain
Please show steps, and explain
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You can tell from the coordinates that the foci are horizontally aligned (both lie on the horizontal line y = 0), so the ellipse is horizontal.
General equation of a horizontal ellipse:
(x - h)²/a² + (y - k)²/b² = 1
with
center (h, k)
a ≥ b
foci (h±c, k), c² = a² - b²
Apply your data to the general equation.
The center is exactly halfway between foci, at (0, 0).
center (h, k) = (0, 0)
h = 0
k = 0
foci (h±c, k) = (±12, 0)
c = 12
b² = a² - c²
The equation becomes
x²/a² + y²/(a² - 144) = 1
(12, 7) is on the ellipse.
12²/a² + 7²/(a² - 12²) = 1
144/a² + 49/(a² - 144) = 1
144(a² - 144) + 49a² = a²(a² - 144)
193a² - 144² = a⁴ - 144a²
a⁴ - 337a² + 144² = 0
∴a² = 256
b² = a² - c² = 112
The equation becomes
x²/256 + y²/112 = 1
When x = 0, y = ±4√7 ≅ ±10.583
When y = 0, x = ±16.
The ellipse intersects the coordinate axes at (0, ±4√7) and (±16, 0).
http://www.flickr.com/photos/dwread/8666…
General equation of a horizontal ellipse:
(x - h)²/a² + (y - k)²/b² = 1
with
center (h, k)
a ≥ b
foci (h±c, k), c² = a² - b²
Apply your data to the general equation.
The center is exactly halfway between foci, at (0, 0).
center (h, k) = (0, 0)
h = 0
k = 0
foci (h±c, k) = (±12, 0)
c = 12
b² = a² - c²
The equation becomes
x²/a² + y²/(a² - 144) = 1
(12, 7) is on the ellipse.
12²/a² + 7²/(a² - 12²) = 1
144/a² + 49/(a² - 144) = 1
144(a² - 144) + 49a² = a²(a² - 144)
193a² - 144² = a⁴ - 144a²
a⁴ - 337a² + 144² = 0
∴a² = 256
b² = a² - c² = 112
The equation becomes
x²/256 + y²/112 = 1
When x = 0, y = ±4√7 ≅ ±10.583
When y = 0, x = ±16.
The ellipse intersects the coordinate axes at (0, ±4√7) and (±16, 0).
http://www.flickr.com/photos/dwread/8666…