Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
Tangents are drawn to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and the circle $x^2 + y^2 = a^2$ at the points where a common ordinate cuts them (on the same side of the $x$-axis). Then the greatest acute angle between these tangents is given by:
$\tan^{-1}\left(\frac{a-b}{2\sqrt{ab}}\right)$
$\tan^{-1}\left(\frac{a+b}{2\sqrt{ab}}\right)$
$\tan^{-1}\left(\frac{2ab}{\sqrt{a-b}}\right)$
$\tan^{-1}\left(\frac{2ab}{\sqrt{a+b}}\right)$
Step-by-Step Solution
Key Concept: The angle between tangents is maximized when the condition $\sqrt{a}\tan\alpha = \sqrt{b}\cot\alpha$ is satisfied.
The tangent to the ellipse at $P(a\cos\alpha, b\sin\alpha)$ is $\frac{x}{a}\cos\alpha + \frac{y}{b}\sin\alpha = 1$, and the tangent to the circle at $Q(a\cos\alpha, a\sin\alpha)$ is $\cos\alpha + \sin\alpha = a$. Using the angle formula between two lines and simplifying, $\tan\theta = \frac{a-b}{2\sqrt{ab}}$ when $\sqrt{a}\tan\alpha = \sqrt{b}\cot\alpha$. Thus, the maximum angle is $\theta_{\text{maximum}} = \tan^{-1}\left(\frac{a-b}{2\sqrt{ab}}\right)$.
Correct Answer: 1