Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 16x2 + y2 = 16

Asked by Pragya Singh | 1 year ago |  114

##### Solution :-

The equation is 16x2 + y2

= 16 or $$\dfrac{x^2}{1}$$ + $$\dfrac{y^2}{16}$$ = 1 or

$$\dfrac{x^2}{1^2}$$ + $$\dfrac{y^2}{4^2}$$ = 1

Here, the denominator of $$\dfrac{y^2}{4^2}$$ is greater than the denominator of $$\dfrac{x^2}{1^2}$$.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation with $$\dfrac{x^2}{b^2}$$+ $$\dfrac{y^2}{a^2}$$ = 1, we get

b =1 and a =4.

c = $$\sqrt{a^2-b^2}$$

$$\sqrt{16-1}$$

$$\sqrt{15}$$

Then,

The coordinates of the foci are (0,$$\sqrt{15}$$) and (0, $$-\sqrt{15}$$).

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

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = $$\dfrac{c}{a}$$$$\dfrac{\sqrt{15}}{4}$$

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

= $$\dfrac{(2×1^2)}{4}$$ = $$\dfrac{2}{4}$$$$\dfrac{1}{2}$$

Answered by Abhisek | 1 year ago

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