Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas 5y2 – 9x2 = 36

Asked by Pragya Singh | 1 year ago |  131

##### Solution :-

The equation is 5y2 – 9x2 = 36

Let us divide the whole equation by 36, we get

$$\dfrac{5y^2}{9}- \dfrac{9x^2}{36}=1$$

$$\dfrac{36}{36}$$

$$\dfrac{y^2}{\dfrac{36}{5}}-\dfrac{x^2}{4}=1$$

On comparing this equation with the standard equation of hyperbola

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

We get a = $$\dfrac{6}{\sqrt{5}}$$ and b = 2,

It is known that, a2 + b2 = c2

So,

c2 = $$\dfrac{36}{5}$$ + 4

c2 = $$\dfrac{56}{5}$$

c = $$\sqrt{\dfrac{56}{5}}$$

$$\dfrac{2\sqrt{14}}{\sqrt{5}}$$

Then,

The coordinates of the foci are

(0, $$\dfrac{2\sqrt{14}}{\sqrt{5}}$$)  and (0, – $$\dfrac{2\sqrt{14}}{\sqrt{5}}$$).

The coordinates of the vertices are

(0,$$\dfrac{6}{\sqrt{5}}$$) and (0, $$- \dfrac{6}{\sqrt{5}}$$).

Eccentricity, e =$$\dfrac{c}{a}$$

$$\dfrac{ \dfrac{2\sqrt{14}}{\sqrt{5}}}{\dfrac{6}{\sqrt{5}}}$$

$$\dfrac{\sqrt{14}}{3}$$

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

$$\dfrac{(2×2^2)}{\dfrac{6}{\sqrt{5}}}$$

$$\dfrac{(2×4)}{\dfrac{6}{\sqrt{5}}}$$

$$\dfrac{4\sqrt{5}}{3}$$

Answered by Abhisek | 1 year ago

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