Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas 9y2 – 4x2 = 36

Asked by Sakshi | 1 year ago |  60

##### Solution :-

Given:

The equation is 9y2 – 4x2 = 36

or $$\dfrac{ y^2}{4} – \dfrac{x^2}{9} = 1$$

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

On comparing this equation with the standard equation of hyperbola $$\dfrac{ y^2}{a^2} – \dfrac{x^2}{b^2} = 1,$$

We get a = 2 and b = 3,

It is known that, a2 + b2 = c2

So,

c2 = 4 + 9

c2 = 13

c = $$\sqrt{13}$$

Then,

The coordinates of the foci are $$(0, \sqrt{13})$$ and $$(0, -\sqrt{13})$$

The coordinates of the vertices are (0, 2) and (0, – 2).

Eccentricity, e = $$\dfrac{c}{a}$$ = $$\sqrt{\dfrac{13}{2}}$$

Length of latus rectum = $$\dfrac{2b^2}{a}$$= $$\dfrac{ (2 × 3^2)}{2}$$

= $$\dfrac{ (2×9)}{2} = \dfrac{18}{2}$$ = 9

Answered by Aaryan | 1 year ago

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