Q

# Solve it, - Two-dimensional Coordinate Geometry - BITSAT

Equation of the hyperbola with eccentricity 3/2 and foci at ($\pm$2, 0) is

• Option 1)

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=\frac{4}{9}$

• Option 2)

$\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{4}{9}$

• Option 3)

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$

• Option 4)

None of these

107 Views

Coordinates of foci -

$\left ( \pm \, ae ,o\right )$

- wherein

For the Hyperbola

$\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1$

$ae=2$

$\Rightarrow a=\frac{4}{3}$

$b^{2}=a^{2}(e^{2}-1)$

$\Rightarrow b^{2}=a^{2}(e^{2}-1)$

$\Rightarrow b^{2}=\frac{16}{9}\left ( \frac{9}{4}-1 \right )$

$=\frac{20}{9}$                                    $\Rightarrow b=\frac{2}{3} \sqrt{5}$

equation $\frac{x^{2}}{\frac{16}{9}}-\frac{y^{2}}{\frac{20}{9}} = 1 \Rightarrow \frac{9x^{2}}{10}-\frac{9y^{2}}{20} = 1$

Option 1)

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=\frac{4}{9}$

This solution is incorrect

Option 2)

$\frac{x^{2}}{9}-\frac{y^{2}}{4}=\frac{4}{9}$

This solution is incorrect

Option 3)

$\frac{x^{2}}{4}-\frac{y^{2}}{9}=1$

This solution is incorrect

Option 4)

None of these

This solution is correct

Exams
Articles
Questions