5.  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.

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

Answers (1)

Given

The equation of ellipse

\frac{x^2}{49} + \frac{y^2}{36} = 1\Rightarrow \frac{x^2}{7^2} + \frac{y^2}{6^2} = 1

As we can see from the equation, the major axis is along X-axis and the minor axis is along Y-axis.

On comparing the given equation with standard equation of ellipse, which is 

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

We get 

a=7 and b=6.

So,

c=\sqrt{a^2-b^2}=\sqrt{7^2-6^2}

c=\sqrt{13}

Hence,

Coordinates of the foci:  

(c,0)\:and\:(-c,0)=(\sqrt{13},0)\:and\:(-\sqrt{13},0)

The vertices:

(a,0)\:and\:(-a,0)=(7,0)\:and\:(-7,0)

The length of major axis:

2a=2(7)=14

The length of minor axis:

2b=2(6)=12

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{13}}{7}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(6)^2}{7}=\frac{72}{7}

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