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Question

A standard hyperbola x2a2−y2b2=1 of eccentricity 'e' is given as below.

Choose the correct options.[S is the focus and dotted line is a directrix.]


A

\)

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B

AB=2a

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C

QS=me

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D

SB=a(e-1)

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Solution

The correct options are
A

\)


B

AB=2a


C

QS=me


D

SB=a(e-1)


This question basically maps the basic parameters of a hyperbola to its graphical form. It also gives an insight into the construction of the

hyperbola as well.

The vertex of the hyperbola are given by the points B (a, 0); A (-a, 0)

CD represents the conjugate axis of the hyperbola.

As you know hyperbola is the locus of points line is a constant, greater than one.

So if Q(m,b) is a points on the hyperbola QSQC=constant=e=QSm

QS=em

S is the focus, given by the point (ae,o),and b(a,0)

SB=aea

=a(e1)


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