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 = \( \dfrac{ 2b^2}{a} =\dfrac{ (2×10^2)}{20}=10\)
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