Question 1185835
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The length of the transverse axis of the equation 100x^2-(y+3)^2=100
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The simplest way to answer the question, without making full analysis of the hyperbola, is THIS.


From the equation, you see that the curve is a hyperbola, whose transverse axis is horizontal  y= - 3, parallel to x-axis.
Its center is the point  (0,-3).

Its imaginary axis is vertical x= 0.  It is the symmetry line between two branches of the hyperbola.


Divide both sides by 100 and write it in the EQUIVALENT form

    x^2 = 1 + 0.01*(y+3)^2


The transverse axis is the shortest distance between the two branches of the hyperbola.

So, we look for minimum value of x^2.


The minimum value of x^2 is when y= -3, and this value is  x^2 = 1 + 0.01*(-3+3)^2 = 1 + 0 = 1,
so  x = +/-1.


Thus the vertices of the hyperbola are  the points (-1,-3)  and  (1,-3).


The distance between them is 2 units horizontally.


So, the length of the transverse axis is 2 units.
</pre>

Solved.


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