1.

Find the eccentricity of the ellipse \(\rm \frac{x^2}{16} + \frac{y^2}{25} = 1\)?1. 12. 2/33. 3/54. 4/5

Answer» 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



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