Q

# 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

103 Views

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

Exams
Articles
Questions