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

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 curvex2 – y2 = 16
Answer» Given equation of the ellipse is x² – y² = 16 Comparing this equation with \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\), we get a² = 16 and b² = 16 ∴ a = 4 and b = 4 (i) Length of major axis = 2a = 2(4) = 8 Length of minor axis = 2b = 2(4) = 8 (ii) We know that e = \(\frac{\sqrt{a^2-b^2}}{a}\) ∴ e =\(\frac{\sqrt{16+16}}{4}\) = √32/4 = 4√2/4 = √2 Co-ordinates of the foci are S(ae, 0) and S'(-ae, 0) i.e., S(4√2), 0) and S'(-4√2, 0) (iii) Equations of the directrices are x = ± a/e i.e., x = ± 4/√2 i.e., x = ± 2√2 (iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(16)}{4} = 8\) (v) Distance between foci = 2ae = 2 (4) (√2) = 8√2 (vi) Distance between directrices = 2a/e = \(\frac {2(4)}{\sqrt2}\)= 4√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 curvex2 – y2 = 16

Answer»

Given equation of the ellipse is x² – y² = 16

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

we get a² = 16 and b² = 16

∴ a = 4 and b = 4

(i) Length of major axis = 2a = 2(4) = 8

Length of minor axis = 2b = 2(4) = 8

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

∴ e =\(\frac{\sqrt{16+16}}{4}\)

= √32/4

= 4√2/4

= √2

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

i.e., S(4√2), 0) and S'(-4√2, 0)

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

i.e., x = ± 4/√2

i.e., x = ± 2√2

(iv) Length of latus rectum = \(\frac{2b^2}{a}= \frac {2(16)}{4} = 8\)

(v) Distance between foci = 2ae = 2 (4) (√2) = 8√2

(vi) Distance between directrices = 2a/e = \(\frac {2(4)}{\sqrt2}\)= 4√2