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 Pragya Singh | 11 months ago |  77

#### 1 Answer

##### Solution :-

The equation is 36x2 + 4y2 = 144 or

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

or $$\dfrac{x^2}{2^2}$$ + $$\dfrac{y^2}{6^2}$$= 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.

On comparing the given equation with

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

= $$\dfrac{(2×2^2)}{6}$$ = $$\dfrac{(2×4)}{6}$$

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

Answered by Abhisek | 11 months ago

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