What are the lengths of the transverse and conjugate axes of the hyperbola given by the equation x^2/62 - y^2/25=1

Question

anonymous · Answer

This looks like Conics.

agent0smith · Answer

Std. form for horizontal hyperbola: $$\frac{ x^2 }{ a^2 } - \frac{ y^2 }{ b^2 }= 1 $$ this should help: http://www.tpub.com/math2/Job%202_files/image619.jpg Transverse axis = 2a Conjugate axis = 2b

anonymous · Answer

it is conics.

anonymous · Answer

so thetransverse axis would be 62 time 2?
an the Conjugat axis 25 times 2? @agent0smith

agent0smith · Answer

Not quite. $$62 = a^2$$$$25 = b^2$$

agent0smith · Answer

Make sense @FlyinSolo_424  ?

anonymous · Answer

wait im confused. Would the transverse just be 62?

anonymous · Answer

@agent0smith

anonymous · Answer

I figured it out!! The transver I would square root it then times it by two and the 62 was ment to be 64. Therefore, Transvers would be 16 and the conjugate 10. Right ? @agent0smith

agent0smith · Answer

Correct! Nice job :)

And i thought that 62 was meant to be a 64...