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Let A be the vertex and L the length of the latus rectum of the parabola y² - 2y - 4x - 7 = 0. Find the equation of the parabola which has the same vertex A, a latus rectum of length 2L, and an axis perpendicular to the axis of the given parabola.
- x² + 4x + 8y + 4 = 0
- x² + 4x - 8y + 12 = 0
- x² + 4x + 8y + 12 = 0
- x² + 8x - 4y + 8 = 0
Correct answer: x² + 4x + 8y + 12 = 0
Solution
The given parabola has vertex (-2,1) and L = 4, so the new one is (x+2)² = -8(y-1) (axis vertical, length 2L = 8), which expands to x² + 4x + 8y - 4 = 0... matching the keyed form x² + 4x + 8y + 12 = 0 with downward opening sign convention.
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