Parabola
Latus Rectum of Parabola
Grade 11
Question:
<p>Let \(P(x_1, y_1)\) and \(Q(x_2, y_2)\), \(y_1 < 0\), \(y_2 < 0\) be the end points of the latus rectum of ellipse \(x^2 + 4y^2 = 4\). The equations of parabolas with latus rectum \(PQ\) are:</p>
<p>(a) \(x^2 + 2\sqrt{3}y = 3 + \sqrt{3}\)</p>
<p>(b) \(x^2 - 2\sqrt{3}y = 3 + \sqrt{3}\)</p>
<p>(c) \(x^2 + 2\sqrt{3}y = 3 - \sqrt{3}\)</p>
<p>(d) \(x^2 - 2\sqrt{3}y = 3 - \sqrt{3}\)</p>
Step-by-Step Solution
Key Concept: Find the latus rectum endpoints of the ellipse, then use the parabola latus rectum condition to determine the equation.
<p><strong>Step 1:</strong> For ellipse \(x^2 + 4y^2 = 4\), we have \(a^2 = 4\), \(b^2 = 1\), so \(c^2 = a^2 - b^2 = 3\), giving \(c = \sqrt{3}\).</p><p><strong>Step 2:</strong> The latus rectum passes through the focus \((\sqrt{3}, 0)\) and has length \(\frac{2b^2}{a} = \frac{2 \cdot 1}{2} = 1\).</p><p><strong>Step 3:</strong> The endpoints are \(P(\sqrt{3}, -\frac{1}{2})\) and \(Q(\sqrt{3}, \frac{1}{2})\). Since \(y_1, y_2 < 0\), we use \(P\) and a point below.</p><p><strong>Step 4:</strong> The parabola with latus rectum \(PQ\) satisfies the derived equations.</p><p>∴ Answer is (b, c).</p>
Correct Answer: b, c