The equation is 16x2 + y2
= 16 or \( \dfrac{x^2}{1}\) + \( \dfrac{y^2}{16}\) = 1 or
\( \dfrac{x^2}{1^2}\) + \( \dfrac{y^2}{4^2}\) = 1
Here, the denominator of \( \dfrac{y^2}{4^2}\) is greater than the denominator of \( \dfrac{x^2}{1^2}\).
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 =1 and a =4.
c = \( \sqrt{a^2-b^2}\)
= \( \sqrt{16-1}\)
= \( \sqrt{15}\)
Then,
The coordinates of the foci are (0,\( \sqrt{15}\)) and (0, \( -\sqrt{15}\)).
The coordinates of the vertices are (0, 4) and (0, -4)
Length of major axis = 2a = 2 (4) = 8
Length of minor axis = 2b = 2 (1) = 2
Eccentricity, e = \( \dfrac{c}{a}\) = \( \dfrac{\sqrt{15}}{4}\)
Length of latus rectum = \( \dfrac{2b^2}{a}\)
= \( \dfrac{(2×1^2)}{4}\) = \( \dfrac{2}{4}\) = \(\dfrac{1}{2}\)
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