NCERT · Class 11 · Mathematics · Conic SectionsThe eccentricity of the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is:
Step-by-Step Solution
For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$, the eccentricity $e$ is given by the formula $e = \sqrt{1 - \frac{b^2}{a^2}}$. Here $a^2 = 25$ and $b^2 = 9$. Thus, $e = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Detailed Options Breakdown
Option : $\frac{3}{5}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
Option 1: $\frac{4}{5}$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: $\frac{9}{25}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
Option 3: $\frac{5}{4}$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of Conic Sections.
💡 Study Guide: This question tests core syllabus concepts from Conic Sections. For formulas, key summaries, and mock exam reference guides, read the full Conic Sections Revision Notes.