यदि किसी दीर्घवृत x^2/(16)+y^(2)/(b^(2))...

यदि किसी दीर्घवृत `x^2/(16)+y^(2)/(b^(2))=1` तथा अतिपरवलय `x^2/(144)-y^2/(81)=1/(25)` की नाभिया सम्पति हो, तब `b^2` का माना होगा।

A

1

B

5

C

7

D

9

Verified by Experts

The correct Answer is:

D

यदि किसी दीर्घवृत x^2/(16)+y^(2)/(b^(2))=1 तथा अतिपरवलय x^2/(144)-y^2/...
Text Solution
|
If the foci of the ellipse (x^2)/(16)+(y^2)/(b^2)=1 and the hyperbola ...
Text Solution
|
If the foci of the ellipse (x^2)/(16)+(y^2)/(b^2)=1 and the hyperbola ...
Text Solution
|
यदि दीर्घवृत्त (x^(2))/(16)+(y^(2))/(b^(2))=1 तथा अतिपरवलय (x^(2))/(14...
Text Solution
|
यदि दीर्घवृत्त (x^(2))/(16)+(y^(2))/(b^(2))=1 व अतिपरवलय (x^(2))/(...
Text Solution
|
If foci of the ellipse (x^(2))/(16) + (y^(2))/(b^(2)) = 1 coincide wit...
Text Solution
|
If the foci of the ellipse x^(2)/16 + y^(2)/b^(2) =1 and the hyperbo...
Text Solution
|
The foci of the ellips (x^(2))/( 16) +(y^(2))/( b^(2) ) =1 and the h...
Text Solution
|
यदि अतिपरवलय (x^(2))/(a^(2))-(y^(2))/(b^(2))=1 की उत्केन्द्रता e तथा...
Text Solution
|