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 Sakshi | 1 year ago |  54

##### Solution :-

Given:

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 Aaryan | 1 year ago

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