Given:
The equation is \( \dfrac{x^2}{36} +\dfrac{ y^2}{16} = 1\)
Here, the denominator of \( \dfrac{x^2}{36}\) is greater than the denominator of \( \dfrac{y^2}{16}\).
So, the major axis is along the x-axis, while the minor axis is along the y-axis.
a = 6 and b = 4.
c = \( \sqrt{(a^2 – b^2)}\)
= \( \sqrt{(36-16)}\)
= \( \sqrt{20}\)
=\( 2\sqrt{5}\)
Then,
The coordinates of the foci are (\( 2\sqrt{5}\), 0) and (\( - 2\sqrt{5}\), 0).
The coordinates of the vertices are (6, 0) and (-6, 0)
Length of major axis = 2a = 2 (6) = 12
Length of minor axis = 2b = 2 (4) = 8
Eccentricity, e = \( \dfrac{c}{a}\) = \(\dfrac{2\sqrt{5}}{6} = \sqrt{\dfrac{5}{3}}\)
Length of latus rectum = \( \dfrac{2b^2}{a} = \dfrac{(2×16)}{6}\)
=\( \dfrac{16}{3} \)
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