Question

Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $9y^2-4x^2=36$

Answer 1

Given hyperbola:$9y^2-4x^2=36$

Dividing it whole by $36$ , $\frac{y^2}{4}-\frac{x^2}{9}=1$

where $b=2,a=3$

Its a conjugate hyperbola so it will have different formula than the normal hyperbola.

Eccentricity $e$$=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{9}{4}}=\frac{1}{2}\sqrt{1}3$

Vertices, $A$ $(0,\pm b)\Rightarrow (0,\pm 2)$

Foci$(0,\pm be)\Rightarrow (0,\pm 2\cdot \frac{1}{2}\sqrt{1}3)\Rightarrow (0,\pm \sqrt{1}3)$

Length of $LR$ $=2\frac{a^2}{b}=2\cdot \frac{9}{4}=\frac{9}{2}$ unit length.