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

$$\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$$

Asked by Pragya Singh | 11 months ago |  61

##### Solution :-

The equation is $$\dfrac{x^2}{4}$$ + $$\dfrac{y^2}{25}$$= 1

Here, the denominator of $$\dfrac{y^2}{25}$$ is greater than the denominator of $$\dfrac{x^2}{4}$$.

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

On comparing the given equation with

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

a = 5 and b = 2.

c = $$\sqrt{ (a^2 – b^2)}$$

$$\sqrt{ 25 – 4}$$

$$\sqrt{21}$$

Then,

The coordinates of the foci are (0,$$\sqrt{21}$$) and (0, $$- \sqrt{21}$$).

The coordinates of the vertices are (0, 5) and (0, -5)

Length of major axis = 2a = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e =$$\dfrac{c}{a}$$$$\dfrac{\sqrt{21}}{5}$$

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

$$\dfrac{(2×2^2)}{5}$$

$$\dfrac{(2×4)}{5}$$

$$\dfrac{8}{5}$$

Answered by Abhisek | 11 months ago

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