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If e and e' are the eccentricities of th...

If `e and e'` are the eccentricities of the hyperbola `x^2/a^2-y^2/b^2=1 and y^2/b^2-x^2/a^2=1,` then the point `(1/e,1/(e'))` lies on the circle (A) `x^2+y^2=1` (B) `x^2+y^2=2` (C) `x^2+y^2=3` (D) `x^2+y^2=4`

A

`x^(2)+y^(2)=1`

B

`x^(2)+y^(2)=2`

C

`x^(2)+y^(2)=3`

D

`x^(2)+y^(2)=4`

Text Solution

Verified by Experts

The correct Answer is:
A
For `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) = 1, e^(2) =1 + (b^(2))/(a^(2)) = (a^(2)+b^(2))/(a^(2))` ltrgt For `(y^(2))/(b^(2)) -(x^(2))/(a^(2)) =1, (e')^(2) =1 + (a^(2))/(b^(2)) =(b^(2)+a^(2))/(b^(2))`
`rArr (1)/(e^(2)) + (1)/((e')^(2))`
`= (a^(2))/(a^(2)+b^(2)) + (b^(2))/(b^(2)+a^(2))`
`= (a^(2)+b^(2))/(a^(2)+b^(2)) =1`
So the point `((1)/(e),(1)/(e))` lie on `x^(2) + y^(2) =1`
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