Find the (i) lengths of the principal axes (ii) co-ordinates of the foci (iii) equations of directrices: x^2/25+ y^2/9

Question

Find the 

(i) lengths of the principal axes 

(ii) co-ordinates of the foci 

(iii) equations of directrices 

(iv) length of the latus rectum 

(v) Distance between foci 

(vi) distance between directrices of the curve

\(\frac {x^2}{25}+ \frac {y^2}{9} = 1\)

x²/25 + y²/9

Answer 1

Given equation of the ellipse is \(\frac {x^2}{25} + \frac {y^2}{9} = 1\)

Comparing this equation with \(\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1\),

we get a² = 25 and b² = 9 

∴ a = 5 and b = 3

Since a > b, 

X-axis is the major axis and Y-axis is the minor axis. 

(i) Length of major axis = 2a = 2(5) = 10 

Length of minor axis = 2b = 2(3) = 6 

∴ Lengths of the principal axes are 10 and 6.

(ii) We know that e = \(\frac{\sqrt{a^2-b^2}}{a}\)

∴ e =\(\frac{\sqrt{25-9}}{5}\)= 4/5

Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0)

i.e., S(5(4/5), 0) and S'(-5(4/5), 0)

i.e., S(4, 0) and S'(-4, 0)

(iii) Equations of the directrices are x = ± a/e

i.e., x = ± \(\frac {5}{\frac{4}{5}}\)

i.e., x = ± 25/4

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(3)^2}{5} = \frac {18}{5}\)

(v) Distance between foci = 2ae = 2 (5) (4/5) = 8

(vi) Distance between directrices = 2a/e = \(\frac {2(5)}{\frac{4}{5}}\)= 25/2