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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 / 4 + y ^2 / 25 = 1

2. 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}{4} + \frac{y^2}{25} =1

Answers (1)

best_answer

Given

The equation of the ellipse

\frac{x^2}{4} + \frac{y^2}{25} =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=5 and b=2.

So,

c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}

c=\sqrt{21}

Hence,

Coordinates of the foci:  

(0,c)\:and\:(0,-c)=(0,\sqrt{21})\:and\:(0,-\sqrt{21})

The vertices:

(0,a)\:and\:(0,-a)=(0,5)\:and\:(0,-5)

The length of the major axis:

2a=2(5)=10

The length of minor axis:

2b=2(2)=4

The eccentricity :

e=\frac{c}{a}=\frac{\sqrt{21}}{6}

The length of the latus rectum:

\frac{2b^2}{a}=\frac{2(2)^2}{5}=\frac{8}{5}

Posted by

Pankaj Sanodiya

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