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 36x2 + 4y2 = 144

Asked by Sakshi | 1 year ago |  68

##### Solution :-

Given:

The equation is 36x2 + 4y2 = 144 or

$$\dfrac{x^2}{4} + \dfrac{y^2}{36 }= 1$$

Here, the denominator of $$\dfrac{ y^2}{6^2}$$ is greater than the denominator of $$\dfrac{ x^2}{2^2}$$

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

b = 2 and a = 6.

c = $$\sqrt{(a^2 – b^2)}$$

= $$\sqrt{(36-4)}$$

$$\sqrt{32}$$

$$4\sqrt{2}$$

Then,

The coordinates of the foci are (0,$$4\sqrt{2}$$) and (0, $$- 4\sqrt{2}$$).

The coordinates of the vertices are (0, 6) and (0, -6)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = $$\dfrac{c}{a}$$ = $$\dfrac{4\sqrt{2}}{6}$$ = $$\dfrac{2\sqrt{2}}{3}$$

Length of latus rectum = $$\dfrac{ 2b^2}{a} =\dfrac{ (2×2^2)}{6}$$

$$\dfrac{4}{3}$$

Answered by Aaryan | 1 year ago

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