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
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