Question
Download Solution PDFIn an ellipse, the distance between its foci is 4 and minor axis is 6.Then its eccentricity is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The general equation of an ellipse is written as: \(\rm \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)
Eccentricity = e =\(\rm \sqrt{1 - \frac{b^{2}}{a^{2}}}\)
Distance between foci of ellipse = 2ae
Calculation:
Given: Distance between its foci is 4
2ae = 4
ae = 2
Given: Minor axis = 6
2b = 6
b = 3
Eccentricity: b2 = a2(1 - e2)
9 = a2 - a2 e2
9 = a2 - 4
9 + 4 = a2
13 = a2
a = \(\rm \sqrt{13}\)
e = \(\rm \frac{c}{a} = \rm \frac{2}{\sqrt{13}}\)
Hence Option 3 is correct.
Last updated on Jul 4, 2025
-> The Indian Coast Guard Navik GD Application Correction Window is open now. Candidates can make the changes in the Application Forms through the link provided on the official website of the Indian Navy.
-> A total of 260 vacancies have been released through the Coast Guard Enrolled Personnel Test (CGEPT) for the 01/2026 and 02/2026 batch.
-> Candidates can apply online from 11th to 25th June 2025.
-> Candidates who have completed their 10+2 with Maths and Physics are eligible for this post.
-> Candidates must go through the Indian Coast Guard Navik GD previous year papers.