Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas  49y2 – 16x2 = 784.

Asked by Pragya Singh | 11 months ago |  114

Solution :-

The equation is 49y2 – 16x2 = 784.

Let us divide the whole equation by 784, we get

$$\dfrac{49y^2}{784}- \dfrac{16x^2}{784}$$

$$\dfrac{784}{784}$$

$$\dfrac{y^2}{16}- \dfrac{x^2}{49}=1$$

On comparing this equation with the standard equation of hyperbola

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

We get a = 4 and b = 7,

It is know that, a2 + b2 = c2

So,

c2 = 16 + 49

c2 = 65

c = $$\sqrt{65}$$

Then,

The coordinates of the foci are (0,$$\sqrt{65}$$) and (0, –$$\sqrt{65}$$).

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

Eccentricity, e = $$\dfrac{c}{a}$$$$\dfrac{\sqrt{65}}{4}$$

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

$$\dfrac{(2×7^2)}{4}$$

$$\dfrac{(2×49)}{4}$$

$$\dfrac{49}{2}$$

Answered by Abhisek | 11 months ago

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