Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
Match the following:
(A) If vertices of a rectangle of maximum area inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are extremities of latus rectum. Then eccentricity of ellipse is
(B) If extremities of diameter of the circle $x^2 + y^2 = 16$ are foci of a ellipse, then eccentricity of the ellipse, if its size is just sufficient to contain the circle, is
(C) If normal at point (6, 2) to the ellipse passes through its nearest focus (5, 2), having centre at (4, 2) then its eccentricity is
(D) If extremities of latus rectum of the parabola $y^2 = 24x$ are foci of ellipse and if ellipse passes through the vertex of the parabola, then its eccentricity is
Step-by-Step Solution
Key Concept: Maximize inscribed rectangle area using parametric ellipse form and relate to focal chord properties.
Part (A): Rectangle with vertices at $(a\cos\theta, b\sin\theta)$ has area $A = 2a\cos\theta \times 2b\sin\theta = 2ab\sin 2\theta$, maximized at $A_{\max} = 2ab$. Rectangle formed by extremities has area $LR = (2ae)(\frac{2b^2}{a}) = 4eb^2$. Setting $2ab = 4eb^2$ gives $e = \frac{1}{\sqrt{2}}$. Part (B): With $2ae = 8$ and $b = 4$, we get $a^2 = 32$, so $e = \frac{1}{\sqrt{2}}$. Part (C): Normal at $P(6,2)$ passes through focus $Q(5,2)$ with $ae = QR = 1$ and $a - ae = 1$. This yields $a = 2$, $b = \sqrt{3}$, and $e = \frac{1}{2}$.
Correct Answer: [A-q] [B-q] [C-s] [D-p]