Find the coordinates of the foci, the vertices the eccentricity an | Filo
filo Logodropdown-logo

Class 11

Math

3D Geometry

Conic Sections

view icon560
like icon150

Find the coordinates of the foci, the vertices the eccentricity and the length of the latus rectum of the hyperbola 

Solution: The given equation is $$\displaystyle 5y^{2}-9x^{2}= 36$$
$$\displaystyle \Rightarrow $$ $$\displaystyle \frac{y^{2}}{\left ( \frac{36}{5} \right )}-\frac{x^{2}}{4}= 1$$
$$\displaystyle \Rightarrow \frac{y^{2}}{\left ( \frac{6}{\sqrt{5}} \right )^{2}} - \frac{x^{2}}{2^{2}} = 1  ...(1)$$
On comparing equation (1) with the standard equation of hyperbola

i.e., $$\displaystyle \frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}= 1$$, 
we obtain $$a = \displaystyle \frac{6}{\sqrt{5}}$$ and  $$b = 2$$
We know that $$\displaystyle a^{2}+ b^{2}= c^{2}, $$ where $$c=ae$$
$$\displaystyle \therefore c^{2}=\frac{36}{5}+ 4 = \frac{56}{5}$$
$$\displaystyle \Rightarrow c = \sqrt{\frac{56}{5}}= \frac{2\sqrt{14}}{\sqrt{5}}$$
Therefore, the coordinates of the foci are $$\displaystyle \left ( 0, \pm \frac{2\sqrt{14}}{\sqrt{5}} \right )$$
The coordinates of the vertices are $$\displaystyle \left ( 0, \pm  \frac{6}{\sqrt{5}} \right )$$
Eccentricity $$e =$$ $$\displaystyle \frac{c}{a}=\frac{\left (

\frac{2\sqrt{14}}{\sqrt{5}} \right )}{\left ( \frac{6}{\sqrt{5}} \right

)}=\frac{\sqrt{14}}{3}$$
Length of latus rectum $$=$$ $$\displaystyle

\frac{2b^{2}}{a}=\frac{2\times 4}{\left ( \frac{6}{\sqrt{5}} \right )}=

\frac{4\sqrt{5}}{3}$$
view icon560
like icon150
filo banner image

Connecting you to a tutor in 60 seconds.

Get answers to your doubts.

playstore logoplaystore logo
Similar Topics
three dimensional geometry
different products of vectors and their geometrical applications
three-dimensional geometry
vector algebra
conic sections