Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

\( \dfrac{x^2}{36}+\dfrac{y^2}{16}=1\)

Asked by Pragya Singh | 11 months ago |  54

1 Answer

Solution :-

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.

On comparing the given equation with 

 \( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1, we get

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 = \( \dfrac{c}{a}\)\( \dfrac{2\sqrt{5}}{6}\)

\( \dfrac{\sqrt{5}}{3}\)

Length of latus rectum = \( \dfrac{2b^2}{a}\)

\( \dfrac{(2×16)}{6}\)

\( \dfrac{16}{3}\)

Answered by Abhisek | 11 months ago

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