Conic Sections
Conic Section
Allen Star Batch
Grade 11

Question:

The extremities of latus rectum of a parabola are $(1, 1)$ and $(1, -1)$, then the equation of the parabola can be :
$y^2 = 2x - 1$
$y^2 = 1 - 2x$
$y^2 = -2x + 3$
$y^2 = 2x - 3$

Step-by-Step Solution

Key Concept: For a parabola with latus rectum endpoints at (1, 1) and (1, -1), the length of latus rectum is 2, giving 4a = 2, so a = 1/2. The focus lies at the midpoint (1, 0), and using the standard form (y - k)² = 4a(x - h) with vertex at (h, k), we get two possible parabolas opening in opposite directions.
Given the extremities of latus rectum are $(1, 1)$ and $(1, -1)$, we have $4a = 2 \Rightarrow a = \frac{1}{2}$. The focus is at $(1, 0)$ and vertices at $(\frac{1}{2}, 0)$ and $(\frac{3}{2}, 0)$. The parabola equations are $y^2 = 2x - 1$ or $y^2 = -2x + 3$, derived from $y^2 = 2(x - \frac{1}{2})$ and $y^2 = -2(x - \frac{3}{2})$ respectively.
Correct Answer: 1,3

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