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

$$\dfrac{x^2}{25}+\dfrac{y^2}{100}=1$$

Asked by Pragya Singh | 1 year ago |  80

##### Solution :-

The equation is $$\dfrac{x^2}{25}$$ + $$\dfrac{y^2}{100}$$ = 1

Here, the denominator of $$\dfrac{y^2}{100}$$ is greater than the denominator of $$\dfrac{x^2}{25}$$.

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

On comparing the given equation with

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

b = 5 and a =10.

c = $$\sqrt{ (a^2 – b^2)}$$

$$\sqrt{ (100– 25)}$$

$$\sqrt{75}$$

$$5 \sqrt{3}$$

Then,

The coordinates of the foci are (0,$$5 \sqrt{3}$$) and (0, $$-5 \sqrt{3}$$).

The coordinates of the vertices are (0,$$\sqrt{10}$$) and (0, $$- \sqrt{10}$$)

Length of major axis = 2a = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

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

$$\dfrac{5\sqrt{3}}{10}$$$$\dfrac{\sqrt{3}}{2}$$

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

= $$\dfrac{(2×5^2)}{10}$$ =$$\dfrac{(2×25)}{10}$$

= 5

Answered by Abhisek | 1 year ago

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