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  • 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.x^2 / 36 + y ^2 / 16 = 1

1. 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}{36} + \frac{y^2}{16} = 1

Answers (1)

best_answer

Given

The equation of the ellipse

\frac{x^2}{36} + \frac{y^2}{16} = 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 the standard equation of an ellipse, which is 

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

We get 

a=6 and b=4.

So,

c=\sqrt{a^2-b^2}=\sqrt{6^2-4^2}

c=\sqrt{20}=2\sqrt{5}

Hence,

Coordinates of the foci:  

(c,0)\:and\:(-c,0)=(2\sqrt{5},0)\:and\:(-2\sqrt{5},0)

The vertices:

(a,0)\:and\:(-a,0)=(6,0)\:and\:(-6,0)

The length of the major axis:

2a=2(6)=12

The length of minor axis:

2b=2(4)=8

The eccentricity :

e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(4)^2}{6}=\frac{32}{6}=\frac{16}{3}

Posted by

Pankaj Sanodiya

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