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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 / 100 + y ^ 2 / 400 = 1

6. 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}{100} + \frac{y^2}{400} =1

Answers (1)

best_answer

Given

The equation of the ellipse

\frac{x^2}{100} + \frac{y^2}{400} =1\Rightarrow \frac{x^2}{10^2} + \frac{y^2}{20^2} =1

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

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

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

We get 

a=20 and b=10.

So,

c=\sqrt{a^2-b^2}=\sqrt{20^2-10^2}

c=\sqrt{300}=10\sqrt{3}

Hence,

Coordinates of the foci:  

(0,c)\:and\:(0,-c)=(0,10\sqrt{3})\:and\:(0,-10\sqrt{3})

The vertices:

(0,a)\:and\:(0,-a)=(0,20)\:and\:(0,-20)

The length of the major axis:

2a=2(20)=40

The length of minor axis:

2b=2(10)=20

The eccentricity :

e=\frac{c}{a}=\frac{10\sqrt{3}}{20}=\frac{\sqrt{3}}{2}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(10)^2}{20}=\frac{200}{20}=10

Posted by

Pankaj Sanodiya

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