 Find the coordinates of the foci, the vertices the eccentricity an | Filo  Class 11

Math

3D Geometry

Conic Sections 560

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 Connecting you to a tutor in 60 seconds.