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Algebra II: Write Equation in Conic section?
1) Parabola with vertex at (3,-2) and focus at (5,-2)
2) Ellipse with vertices at (-5,1) and (-1,1) and co vertices at (-3,2) and (-3,0).
3) Hyperbola with vertices at (8,-4) and (8,4) and foci at (8,-6) and (8,6)
2 Answers
- DWReadLv 78 years ago
Determine orientation.
(3, -2) and (5, -2) are horizontally aligned (both lie on the line y = -2), so the parabola is horizontal.
The focus lies to the right of the vertex, so the parabola opens to the right.
Equation of a right-opening parabola:
x = a(y - k)² + h
with
vertex (h, k)
focus (h+p, k), where p = 1/(4a)
- davisonLv 45 years ago
you will possibly desire to finished the sq. for the 72y + 36y^2, yet first divide the completed factor via 4 to get: x^2 + 9y^2 + 18y = 27 element the 9y^2 + 18y = 9(y^2+2y) = 9(y+a million)^2 - 9, and plug it back in x^2 + 9(y+a million)^2 - 9 = 27 <=> x^2 + 9(y+a million)^2 = 36, now divide via 36, to get a million on the superb area x^2/36 + (y+a million)^2/4 = a million <=> x^2/6^2 + (y+a million)^2/2^2 = a million wish this helped
