Find the (i) lengths of the principal axes (ii) co-ordinates of the foci (iii) equations of directrices :x^2/144 - y^2/25 =1

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

\(\frac {x^2}{144} - \frac {y^2}{25} = 1\)

x²/144 - y²/25 =1

Answer 1

Given equation of the ellipse is \(\frac {x^2}{144} - \frac {y^2}{25} = 1\)

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

we get a² = 144 and b² = 25

∴ a = 12 and b = 5

(i) Length of major axis = 2a = 2(12) = 24

Length of minor axis = 2b = 2(5) = 10

∴ Lengths of the principal axes are 24 and 10.

(ii) b² = a² (e² – 1)

25 = 144 (e² – 1)

25/144 = e² – 1

e² = 1 + 25/144

e² = 169/144

e = 13/12 …….[∵ e > 1]

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

i.e., S(12(13/12), 0) and S'(-12(13/12), 0)

i.e., S(13, 0) and S'(-13, 0)

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

i.e., x = ± \(\frac {12}{\frac{13}{12}}\)

i.e., x = ± 144/13

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(25)}{12} = \frac {25}{6}\)

(v) Distance between foci = 2ae = 2 (12) (13/12) = 26

(vi) Distance between directrices = 2a/e = \(\frac {2(12)}{\frac{13}{12}}\)= 288/13