<p>Let an ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) and a parabola \(x^2 = 2y\) intersecting each other above x-axis at P and Q. Tangents to ellipse at P and Q intersect y-axis at R and tangents to parabola at P and Q intersect y-axis at S, then the area of quadrilateral PRQS is \(\lambda\sqrt{3}\), then \([\lambda]\) = (where [.] denote greatest integer function)</p>
Step-by-Step Solution
Key Concept: Find intersection points by solving the system of ellipse and parabola equations. Then find tangent lines at these points and their y-intercepts to determine the quadrilateral vertices.
<p><strong>Solution:</strong> First, find the intersection points P and Q of the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) and parabola \(x^2 = 2y\).</p><p>Substituting \(x^2 = 2y\): \(\frac{2y}{4} + \frac{y^2}{9} = 1\) gives \(\frac{y}{2} + \frac{y^2}{9} = 1\).</p><p>This simplifies to: \(9y + 2y^2 = 18\), or \(2y^2 + 9y - 18 = 0\).</p><p>Using quadratic formula: \(y = \frac{-9 \pm \sqrt{81 + 144}}{4} = \frac{-9 \pm 15}{4}\).</p><p>So \(y = \frac{3}{2}\) or \(y = -6\). Since above x-axis, \(y = \frac{3}{2}\).</p><p>From \(x^2 = 2y = 3\): \(x = \pm\sqrt{3}\).</p><p>So \(P = (\sqrt{3}, \frac{3}{2})\) and \(Q = (-\sqrt{3}, \frac{3}{2})\).</p><p>For ellipse tangent at P: \(\frac{\sqrt{3}x}{4} + \frac{3y}{18} = 1\), simplifying: \(\frac{\sqrt{3}x}{4} + \frac{y}{6} = 1\).</p><p>At y-axis (\(x=0\)): \(R = (0, 6)\).</p><p>For parabola \(x^2 = 2y\), tangent at P is: \(\sqrt{3}x = y + \frac{3}{2}\).</p><p>At y-axis: \(S = (0, -\frac{3}{2})\).</p><p>Area of PRQS = \(\frac{1}{2}|x_P - x_Q| \cdot |y_R - y_S| = \frac{1}{2} \cdot 2\sqrt{3} \cdot \frac{15}{2} = \frac{15\sqrt{3}}{2}\).</p><p>So \(\lambda = \frac{15}{2} = 7.5\), thus \([\lambda] = 7\).</p>
Correct Answer: 7