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

$$\dfrac{x^2}{49}+\dfrac{y^2}{36}=1$$

Asked by Pragya Singh | 1 year ago |  111

##### Solution :-

The equation is $$\dfrac{x^2}{49}$$ + $$\dfrac{y^2}{36}$$ = 1

Here, the denominator of $$\dfrac{x^2}{49}$$ is greater than the denominator of $$\dfrac{y^2}{36}$$.

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

b = 6 and a =7

c = $$\sqrt{ (a^2 – b^2)}$$

$$\sqrt{ (49– 36)}$$

$$\sqrt{ 13}$$

Then,

The coordinates of the foci are ($$\sqrt{ 13}$$, 0) and ($$- \sqrt{ 13}$$, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = $$\dfrac{c}{a}$$$$\dfrac{\sqrt{13}}{7}$$

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

= $$\dfrac{(2×6^2)}{6}$$ = $$\dfrac{(2×36)}{7}$$

$$\dfrac{72}{7}$$

Answered by Abhisek | 1 year ago

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