Find the eccentricity of the ellipse \(\rm \frac{x^2}{16} + \frac{y^2}{25} = 1\)?
1. 1
2. 2/3
3. 3/5
4. 4/5
1. 1
2. 2/3
3. 3/5
4. 4/5
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Correct Answer – Option 3 : 3/5
Concept:
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