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Q.

e and e¹ are the eccentricities of the hyperbolas 16x² – 9y² = 144 and 9x² – 16y² = - 144 then e – e¹ =

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a

1

b

0

c

3

d

5

answer is B.

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Detailed Solution

$e = \sqrt{\frac{a^{2} {+b}^{2}}{a^{2}}} {, e}^{1} = \sqrt{\frac{a^{2} {+ b}^{2}}{b^{2}}}$

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e and e1 are the eccentricities of the hyperbolas 16x2 – 9y2 = 144 and 9x2 – 16y2 = - 144 then e – e1 =