The locus of the point of intersection of the straight lines,

tx−2y−3t=0

x−2ty+3=0  (t \small \epsilon R), is :

 

  • Option 1)

     an ellipse with eccentricity \frac{2}{\sqrt{5}}

  • Option 2)

     an ellipse with the length of major axis 6

     

  • Option 3)

    a hyperbola with eccentricity \sqrt{5}

  • Option 4)

     a hyperbola with the length of conjugate axis 3

     

 

Answers (2)
V Vakul

As we  learnt in

Locus -

Path followed by a point p(x,y) under given condition (s).

- wherein

It is satisfied by all the points (x,y) on the locus.

 

 

Conjugate axis -

The line through the centre and perpendicular to transverse axis.

- wherein

 

 tx-2y-3t=0

x-2ty+3=0 (t\epsilon R)

On solving, x=\frac{-3(t^{2}+1)}{t^{2}-1}

y=\frac{3t}{(t^{2}-1)}

x^{2}-4y^{2}=\frac{9}{(t^{2}-1^{2})}\left [ (t^{2}+1)^{2} -t^{2}\right ]

x^{2}-4y^{2}=9

\frac{x^{2}}{9}-\frac{y^{2}}{(9/4)}=1

Here a=3, b=\frac{3}{2}

Length of conjugrate axis

=2b=3


Option 1)

 an ellipse with eccentricity \frac{2}{\sqrt{5}}

This option is incorrect

Option 2)

 an ellipse with the length of major axis 6

 

This option is incorrect

Option 3)

a hyperbola with eccentricity \sqrt{5}

This option is incorrect

Option 4)

 a hyperbola with the length of conjugate axis 3

 

This option is correct

N neha

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