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| 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\(\frac {x^2}{25}+ \frac {y^2}{9} = 1\)x2/25 + y2/9 | 
| Answer» 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 | |