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 Aaryan | 1 year ago |  38

##### Solution :-

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 = $$\dfrac{2b^2}{a} = \dfrac{(2×16)}{6}$$

=$$\dfrac{16}{3}$$

Answered by Aaryan | 1 year ago

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