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

$$\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$$

Asked by Pragya Singh | 1 year ago |  86

Solution :-

The equation is $$\dfrac{x^2}{16}- \dfrac{y^2}{9}=1$$

or $$\dfrac{x^2}{4^2}- \dfrac{y^2}{3^2}=1$$

On comparing this equation with the standard equation of hyperbola

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

We get a = 4 and b = 3,

It is known that, a2 + b2 = c2

So,

c2 = 42 + 32

= $$\sqrt{25}$$

c = 5

Then,

The coordinates of the foci are (±5, 0).

The coordinates of the vertices are (±4, 0).

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

$$\dfrac{5}{4}$$

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

$$\dfrac{(2×3^2)}{4}$$

$$\dfrac{(2×9)}{4}$$

$$\dfrac{18}{4}$$

$$\dfrac{9}{2}$$

Answered by Abhisek | 1 year ago

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