The equation is \( \dfrac{x^2}{25}\) + \( \dfrac{y^2}{100}\) = 1
Here, the denominator of \( \dfrac{y^2}{100}\) is greater than the denominator of \( \dfrac{x^2}{25}\).
So, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation with
\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1, we get
b = 5 and a =10.
c = \( \sqrt{ (a^2 – b^2)}\)
= \( \sqrt{ (100– 25)}\)
= \( \sqrt{75}\)
= \(5 \sqrt{3}\)
Then,
The coordinates of the foci are (0,\( 5 \sqrt{3}\)) and (0, \( -5 \sqrt{3}\)).
The coordinates of the vertices are (0,\( \sqrt{10}\)) and (0, \(- \sqrt{10}\))
Length of major axis = 2a = 2 (10) = 20
Length of minor axis = 2b = 2 (5) = 10
Eccentricity, e =\( \dfrac{c}{a}\)
= \( \dfrac{5\sqrt{3}}{10}\) = \( \dfrac{\sqrt{3}}{2}\)
Length of latus rectum =\( \dfrac{2b^2}{a}\)
= \( \dfrac{(2×5^2)}{10}\) =\( \dfrac{(2×25)}{10}\)
= 5
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