Find the (i) lengths of the principal axes (ii) co-ordinates of the

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 ellipse:\(\frac {x^2}{25} + \frac {y^2}{9}=1\)x2/25 + y2/9 = 1
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 = \(\frac {\sqrt{a^2-b^2}}{a}\) = \(\frac {\sqrt{25-9}}{5}\) = \(\frac {\sqrt16}{5}\) = 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(5(4, 0) and S' (-4, 0) (iii) Equations of the directrices are x = ± a/e = ± 5/4 = ± 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

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 ellipse:\(\frac {x^2}{25} + \frac {y^2}{9}=1\)x2/25 + y2/9 = 1

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 = \(\frac {\sqrt{a^2-b^2}}{a}\)

= \(\frac {\sqrt{25-9}}{5}\)

= \(\frac {\sqrt16}{5}\)

= 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(5(4, 0) and S' (-4, 0)

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

= ± 5/4

= ± 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