<p>A circle concentric to the ellipse \(\frac{4x^2}{289} + \frac{y^2}{l} = 1\) \(\left(l < \frac{17}{2}\right)\) passes through foci \(S_1, S_2\) and cuts ellipse at point P. If area of \(\triangle PS_1S_2\) is 30 sq. units then find \(\frac{S_1S_2}{13}\).</p>
Step-by-Step Solution
Key Concept: A circle centered at the origin passing through both foci has radius equal to the distance from center to each focus. The area condition constrains the position of P on the ellipse.
<p><strong>Solution:</strong> From the ellipse equation \(\frac{4x^2}{289} + \frac{y^2}{l} = 1\), we have \(a^2 = \frac{289}{4}\), so \(a = \frac{17}{2}\).</p><p>The circle passes through the foci, so its radius is \(c\) where \(c^2 = a^2 - b^2 = \frac{289}{4} - l\).</p><p>Since P lies on both the circle and ellipse: \(x_P^2 + y_P^2 = c^2\).</p><p>Also from ellipse: \(\frac{4x_P^2}{289} + \frac{y_P^2}{l} = 1\).</p><p>The distance \(S_1S_2 = 2c\).</p><p>Area of \(\triangle PS_1S_2 = \frac{1}{2} \cdot 2c \cdot h = 30\), where \(h\) is the perpendicular distance from P to \(S_1S_2\).</p><p>This gives \(c \cdot h = 30\).</p><p>From the geometry and constraint \(l < \frac{17}{2}\), we find \(c = 13/2\), so \(S_1S_2 = 13\).</p><p>Therefore, \(\frac{S_1S_2}{13} = 1\).</p>
Correct Answer: 1