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If e1 is the eccentricity of the ellipse...

If `e_1` is the eccentricity of the ellipse `x^2/16+y^2/25=1 and e_2` is the eccentricity of the hyperbola passing through the foci of the ellipse and `e_1 e_2=1`, then equation of the hyperbola is

A

`(x^(2))/(9)-(y^(2))/(16)=1`

B

`(x^(2))/(16)-(y^(2))/(9)= -1`

C

`(x^(2))/(9)-(y^(2))/(25)= 1`

D

None of these

Text Solution

Verified by Experts

The eccentricity of `(x^(2))/(16)+(y^(2))/(25)=1` is
`e_(1)=sqrt(1-(16)/(25))=(3)/(5)`
` therefore e_(2)=(5)/(3) " "[ because e_(1)e_(2)=1]`
`rArr " Foci of ellipse " (0, pm3)`
`rArr " Equation of hyperbola is " (x^(2))/(16)-(y^(2))/(9)= -1`.
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