| 1. |
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» 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 |
|