Question

Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity, and the length of the latus rectum of the ellipse \(\dfrac{x^2}{100}+\dfrac{y^2}{400}=1\)

Answer 1

The given equation is \(\dfrac{x^2}{100}+\dfrac{y^2}{400}=1\) or  \(\dfrac{x^2}{10^2}+\dfrac{y^2}{20^2}=1\)

Here, the denominator of \(\dfrac{y^2}{400}\)  is greater than the denominator of \(\dfrac{x^2}{100}.\)
Therefore, the major axis is along the \(y-axis\), while the minor axis is along the\( x-axis.\)
On comparing the given equation with \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\), we obtain \(b = 10\) and \(a = 20.\)
\(∴ c = √(a^2 – b^2)\)

\(= √(400-100)\)

\(= √300\)

\(= 10√3\)

Therefore, 
The coordinates of the foci are \((0, ±10√3). \)
The coordinates of the vertices are \((0, ±20) \)
Length of major axis =\( 2a= 40\)
Length of the minor axis = \(2b = 2 (10) = 20\)
Eccentricity, \(e = c/a = 10√3/20 = √3/2\)
Length of latus rectum = \(\dfrac{2b^2}{a} = \dfrac{(2×10^2)}{20} =\dfrac{(2×100)}{20} = 10\)