Find the coordinates of the foci, the vertices the eccentricity an | Filo

Class 11

Math

3D Geometry

Conic Sections

560
150

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 $$=$$ $$\displaystyle \frac{2b^{2}}{a}=\frac{2\times 4}{\left ( \frac{6}{\sqrt{5}} \right )}= \frac{4\sqrt{5}}{3}$$
560
150

Connecting you to a tutor in 60 seconds.

Get answers to your doubts.

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