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```				   IF PQ IS A DOUBLE ORDINATE OF THE HYPERBOLA x2/a2-y2/b2=1 SUCH THAT OPQ IS AN EQUILATERAL TRIANGLE O BEING THE CENTRE OF THE HYPERBOLA.PROVE THAT THE ECCENTRICITY e>2/√3.
```

6 years ago

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```										Dear student,
take this hint:
x-h)2/a2 +(y-k)2/b2 =1  is the equation of the hyperbola with centre (h,k)
x2/a2-y2/b2 =1  is the equation with centre (0,0)
The  standard things here are horizontal axis is the axis passing through  two foci and centre and vertical axis is one perpendicular to horizontal  one.  There are two asymptotes to the hyperbola such that the hyperbola  lies bounded by these two on four sides.  The slope of the asymptotes  are normally b/a with positive and negative sign and the asymptotes  definitely pass through the centre.
Foci  are two points lying inside hyperbola such that the difference of  distance between the two foci always remain constant for any point in  hyperbola.
Directrices are two lines  parallel to minor axis such that the distance from the directrix to any  point is equal to the distance from the focus to the point.
Hence the above para gives a brief outline about a hyperbola.
Eccentricity:
The  eccentricity is a crucial point in hyperbola and the eccentricity  decides the shape of the hyperbola.  Eccentricity denoted by e is always  >1 in a hyperbola whereas in ellipse it lies between 0 and 1.
Definition:
The eccentricity  of an ellipse is defined as the ratio of the distance between any point  in the ellipse and the focus and distance between point and a fixed  line called directrix.

e = distance between P and F/distance between P and line D (Directrix)
Formula for eccentricity :
Formula for calculating e is
b2 = (a2)(e2-1)    So e2- 1 = b2/a2     or e2 = b2/a2 +1
e = √(b2/a2+1)
```
5 years ago

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