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 | 2 years ago |  117

1 Answer

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 = \( \sqrt{ (a^2 – b^2)}\)

\( \sqrt{ (400– 100)}\)

\( \sqrt{ 300}\)

\(10 \sqrt{3}\)


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 | 2 years ago

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