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

$$\dfrac{y^2}{9}-\dfrac{x^2}{27}=1$$

Asked by Pragya Singh | 11 months ago |  80

##### Solution :-

The equation is $$\dfrac{y^2}{9}- \dfrac{x^2}{27}=1$$

or $$\dfrac{y^2}{3^2}- \dfrac{x^2}{27^2}=1$$

On comparing this equation with the standard equation of hyperbola

$$\dfrac{x^2}{a^2}$$ +$$\dfrac{y^2}{b^2}$$ =1 ,

We get a = 3 and b =$$\sqrt{27}$$,

It is known that, a2 + b2 = c2

So,

c2 = 32 + ($$\sqrt{27}$$)2

= 9 + 27

c2 = 36

c = $$\sqrt{36}$$

= 6

Then,

The coordinates of the foci are (0, 6) and (0, -6).

The coordinates of the vertices are (0, 3) and (0, – 3).

Eccentricity, e = $$\dfrac{c}{a}$$

=$$\dfrac{6}{3}$$ = 2

Length of latus rectum = $$\dfrac{2b^2}{a}$$

$$\dfrac{(2×27)}{3}$$

$$\dfrac{54}{3}$$ = 18

Answered by Abhisek | 11 months ago

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