Question

Qns5. Find the equation of the hyperbola with vertices (0,±√11/2) and eccentricity 6/√11.
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Answer 1

Given data:

The vertices of the hyperbola are $(0,\pm \sqrt{\frac{11}{2}})$

and eccentricity, $e=\frac{6}{\sqrt{11}}$

To find: the eqn. of the hyperbola

Step-by-step explanation:

Here vertices are $(0,\pm \sqrt{\frac{11}{2}})$

$\Rightarrow (0,\pm b)\equiv (0,\pm \sqrt{\frac{11}{2}})$

Then $b=\sqrt{\frac{11}{2}}$

Here, $e=\frac{6}{\sqrt{11}}$

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

$\Rightarrow a^{2}=\frac{11}{2}\:(\frac{36}{11}-1)$

$\Rightarrow a^{2}=\frac{25}{2}$

Thus the equation of the hyperbola is

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

$\Rightarrow \frac{x^{2}}{\frac{25}{2}}-\frac{y^{2}}{\frac{11}{2}}=-1$

$\Rightarrow 22x^{2}-50y^{2}=275$

Answer:

The required hyperbola is

$\quad\quad 22x^{2}-50y^{2}=275$

NOTE:

If the vertices are $(0,\pm \frac{\sqrt{11}}{2}),$ kindly approach with the similar method as solved above.