Question

Find the coordinate of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

1. \(\frac{x^2}{4} + \frac{y^2}{25} = 1\)

2. \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)

Answer 1

1. Since 25 > 4 the standard equation of the ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\)

⇒ a² = 25; b² = 4

c² = a² – b² = 25 – 4 = 21 

⇒ c = √21

Coordinate of foci are (0, ±√21)

Coordinate of vertex are (0, ±5)

Length of major axis = 2a = 2 × 5 = 10

Length of minor axis = 2b = 2 × 2 = 4

Eccentricity = e = \(\frac{c}{a}\) =\(\frac{\sqrt{21}}{5}\)

Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5}.\)

2. Since 16 > 9 the standard equation of the ellipse is \(\frac{x^2}{a} + \frac{y^2}{b^2} = 1\)

⇒ a² = 16; b² = 9

c² = a² – b² = 16 – 9 = 7 

⇒ c = √7

Coordinate of foci are (±√7, 0)

Coordinate of vertex are (±4, 0)

Length of major axis = 2a = 2 × 4 = 8

Length of minor axis = 2b = 2 × 3 = 6

Eccentricity = e = \(\frac{c}{a}\)=\(\frac{\sqrt{7}}{4}\)

Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}.\)