Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
A normal is drawn to the ellipse $\frac{x^2}{(a^2+2a+1)^2} + \frac{y^2}{(a^2+1)^2} = 1$, $a > 0$ whose centre is at $O$. If maximum radius of the circle, centered at the origin and touching the normal, is $5$ then the positive value of $'a'$ is:....
Step-by-Step Solution
Key Concept: The maximum radius of a circle centered at origin touching any normal to the ellipse equals the minimum of the two semi-axes lengths.
Let $A = (a^2+2a+1)^2 = (a+1)^4$ and $B = (a^2+1)^2$. The ellipse is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. For a normal at point $(A\cos\theta, B\sin\theta)$, the distance from origin to the normal line is given by $d = \frac{AB}{\sqrt{A^2\sin^2\theta + B^2\cos^2\theta}}$. The maximum distance occurs when the denominator is minimized, giving $d_{max} = \min(A, B)$. Since we need $\max(d) = 5$ and $A = (a+1)^4 > B = (a^2+1)^2$ for $a > 0$, we have $B = 5$. Thus $(a^2+1)^2 = 25$, so $a^2+1 = 5$ (taking positive root), yielding $a^2 = 4$, hence $a = 2$.
Correct Answer: 2