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 Aaryan | 1 year ago |  68

#### 1 Answer

##### Solution :-

Given:

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.

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 = $$\dfrac{c}{a}$$= $$\dfrac{10\sqrt{3}}{20}=\sqrt{\dfrac{3}{2}}$$

Length of latus rectum = $$\dfrac{ 2b^2}{a} =\dfrac{ (2×10^2)}{20}=10$$

Answered by Aaryan | 1 year ago

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