Conic Sections
Conic Section
Allen Star Batch
Grade 11
Question:
Which of the following is/are true?
There are infinite positive integral values of $a$ for which $(13x-1)^2 + (13y-2)^2 = \left(\frac{5x+12y-1}{a}\right)^2$ represents an ellipse
The minimum distance of a point $(1, 2)$ from the ellipse $4x^2 + 9y^2 + 8x - 36y + 4 = 0$ is 1
If from a point $P(0, a)$ (P is not the origin) two normals other than axes are drawn to the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, then $|a| < \frac{9}{4}$
If the length of latus rectum of an ellipse is one-third of its major axis, then its eccentricity is equal to $\frac{1}{\sqrt{3}}$
Step-by-Step Solution
Key Concept: Multiple properties of ellipses—eccentricity, distance formulas, and normal equations—are applied to verify different conditions.
For option (A), the equation $(x-\frac{1}{13})^2 + (y-\frac{2}{13})^2 = \frac{1}{a^2}(\frac{5x+12y-1}{13})^2$ represents an ellipse when $\frac{1}{a^2} 1$. For option (B), rewriting $4x^2 + 8x + 9y^2 - 36y = -4$ gives $\frac{(x+1)^2}{9} + \frac{(y-2)^2}{4} = 1$ with center $(-1,2)$ and minimum distance to $(1,2)$ is $1$. For option (C), the normal at $P(θ)$ gives $|a| < \frac{9}{4}$. For option (D), using $\frac{2b^2}{a} = \frac{2a}{3}$ yields $b^2 = a^2(1-e^2) = 3(1-e^2)$, solving gives $e = \sqrt{\frac{2}{3}}$.
Correct Answer: 1,2,3