<p>A normal is drawn to the ellipse <span class="math">\(\frac{x^2}{(a^2 + 2a + 2)^2} + \frac{y^2}{(a^2 + 1)^2} = 1\)</span> whose centre is at O. If maximum radius of the circle, centered at the origin and touching the normal, is 5 then find the positive value of 'a'.</p>
Step-by-Step Solution
Key Concept: The maximum distance from the origin to a normal of an ellipse equals the semi-major axis. The radius of the largest circle centered at origin that touches any normal equals this maximum distance, which we set equal to 5.
<p><strong>Step 1:</strong> Set up the ellipse equation with A² = (a² + 2a + 2)² and B² = (a² + 1)². The ellipse is: $\frac{x^2}{(a^2+2a+2)^2} + \frac{y^2}{(a^2+1)^2} = 1$</p><p><strong>Step 2:</strong> For a point (x₀, y₀) on the ellipse, the normal line equation is: $\frac{A^2 x}{x_0} - \frac{B^2 y}{y_0} = A^2 - B^2$</p><p><strong>Step 3:</strong> The distance from origin O(0,0) to this normal is: $d = \frac{|A^2 - B^2|}{\sqrt{\frac{A^4}{x_0^2} + \frac{B^4}{y_0^2}}}$</p><p><strong>Step 4:</strong> The maximum distance from origin to any normal of the ellipse equals the semi-major axis A (when A > B). Therefore: $\text{max distance} = A = (a^2 + 2a + 2)$</p><p><strong>Step 5:</strong> The maximum radius of a circle centered at origin touching the normal equals this maximum distance: $a^2 + 2a + 2 = 5$</p><p><strong>Step 6:</strong> Solve $a^2 + 2a + 2 = 5$: $a^2 + 2a - 3 = 0$. Factoring: $(a + 3)(a - 1) = 0$. So $a = -3$ or $a = 1$.</p><p><strong>Step 7:</strong> Verify both values give A > B: For a = 1: A² = (1+2+2)² = 25 and B² = (1+1)² = 4, so A = 5 > B = 2 ✓. For a = -3: A² = (9-6+2)² = 25 and B² = (9+1)² = 100, so A = 5 < B = 10 ✗</p><p><strong>Step 8:</strong> The maximum distance applies when A is the semi-major axis. For a = 1, we have A = 5 which matches our requirement. For a = -3, B becomes the semi-major axis but equals 10, not 5.</p><p><strong>∴ Answer: 2</strong></p>
Correct Answer: 2