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 Sakshi | 1 year ago |  48

Solution :-

Given:

The equation is 5y2 – 9x2 = 36

Let us divide the whole equation by 36, we get

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

$$\dfrac{ y^2}{(\dfrac{36}{5})} – \dfrac{x^2}{4} = 1$$

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

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 =$$\dfrac{c}{a}$$ = ($$\dfrac{2\sqrt{14}}{\sqrt{5}}$$)$$\dfrac{6}{\sqrt{5}}$$ = $$\dfrac{14}{\sqrt{3}}$$

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 Aaryan | 1 year ago

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