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

$$\dfrac{x^2}{100}+\dfrac{y^2}{400}=1$$

Asked by Pragya Singh | 1 year ago |  90

##### Solution :-

The equation is $$\dfrac{x^2}{100}$$ + $$\dfrac{y^2}{400}$$ = 1

Here, the denominator of $$\dfrac{y^2}{400}$$ is greater than the denominator of $$\dfrac{x^2}{100}$$.

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 = 10 and a =20.

c = $$\sqrt{ (a^2 – b^2)}$$

$$\sqrt{ (400– 100)}$$

$$\sqrt{ 300}$$

$$10 \sqrt{3}$$

Then,

The coordinates of the foci are (0,$$10 \sqrt{3}$$) and (0, $$- 10 \sqrt{3}$$).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

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

$$\dfrac{10\sqrt{3}}{20}$$$$\dfrac{\sqrt{3}}{2}$$

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

= $$\dfrac{(2×10^2)}{20}$$ = $$\dfrac{(2×100)}{20}$$ = 10

Answered by Abhisek | 1 year ago

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