Question
Download Solution PDFFind the eccentricity of the ellipse \(\rm \frac{x^2}{16} + \frac{y^2}{25} = 1\)?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Ellipse:
|
Equation |
\(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (a > b) |
\(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (a < b) |
|
Equation of Major axis |
y = 0 |
x = 0 |
|
Equation of Minor axis |
x = 0 |
y = 0 |
|
Length of Major axis |
2a |
2b |
|
Length of Minor axis |
2b |
2a |
|
Vertices |
(± a, 0) |
(0, ± b) |
|
Focus |
(± ae, 0) |
(0, ± be) |
|
Directrix |
x = ± a/e |
y = ± b/e |
|
Center |
(0, 0) |
(0, 0) |
|
Eccentricity |
\(\rm \sqrt{1-\frac{b^2}{a^2}}\) |
\(\rm \sqrt{1-\frac{a^2}{b^2}}\) |
Calculation:
Given: \(\rm \frac{x^2}{16} + \frac{y^2}{25} = 1\)
Compare with the standard equation \(\rm \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
So, a2 = 16 and b2 = 25 ⇔ a = 4 and b = 5 (b > a)
So, eccentricity = \(\rm \sqrt{1-\frac{a^2}{b^2}}\)
= \(\sqrt{1-\frac{16}{25}}\)
= 3/5
Last updated on Jun 11, 2025
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