Given:
The equation is 16x2 + y2 = 16 or
\( \dfrac{x^2}{1} + \dfrac{y^2}{16 }= 1\)
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.
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 = \( \dfrac{c}{a}\) = \( \sqrt{\dfrac{15}{4}}\)
Length of latus rectum = \(\dfrac{ 2b^2}{a} =\dfrac{ (2×1^2)}{4}\)
= \( \dfrac{2}{4}\)= \( \dfrac{1}{2}\)
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