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  • Give answer! - Two-dimensional Coordinate Geometry - BITSAT

Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity \sqrt{3} 

  • Option 1)

    7x2 - 2y2 + 12xy - 2x + 14y - 22 = 0

  • Option 2)

    7x2 - 2y2 + 2xy - 2x + 14y - 22 = 0

  • Option 3)

    7x2 - 2y2 + xy - 14x + 2y - 22 = 0

  • Option 4)

    none of the above

 

Answers (1)

best_answer

 

Eccentricity -

The ratio of a distance of point from focus to distance from fixed line.

- wherein

  It is donated by e.

 

 &

Directrix -

The fixed straight line of a conic section.

- wherein

 

 &

Focus -

The fixed point of a conic section

- wherein

 

 \frac{distance\:\: from\:\: focus}{distance\:\: from\:\: line} = \sqrt{3}

\Rightarrow \frac{\sqrt{(x-1)^{2}+(4-2)^{2}}}{\left | 2x+y-1 \right |} =\sqrt{3}

\Rightarrow 5(x-1)^{2}+5(4-2)^{2}=3 (2x+4-1)^{2}

\Rightarrow 5x^{2}+5y^{2}-10x-20y+5+20

=12x^{2}+3y^{2}+3+12xy-6y-12x

\Rightarrow 7x^{2}-2y^{2}+12xy-2x+14y-22=0


Option 1)

7x2 - 2y2 + 12xy - 2x + 14y - 22 = 0

This solution is correct

Option 2)

7x2 - 2y2 + 2xy - 2x + 14y - 22 = 0

This solution is incorrect

Option 3)

7x2 - 2y2 + xy - 14x + 2y - 22 = 0

This solution is incorrect

Option 4)

none of the above

This solution is incorrect

Posted by

prateek

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