Question

Find the lengths of the major and minor axes; coordinates of the vertices and the foci; the eccentricity and length of the latus rectum of the ellipse.

$4x^2+9y^2=144.$

Answer 1

The given equation may be written as,

$\frac{x^2}{36}+\frac{y^2}{16}=1.$

This is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where $a^2>b^2.$

So, it is an equation of a horizontal ellipse.

Now, $(a^2=36and{b}^2=16)\Rightarrow (a=6andb=4).$

$\therefore c=\sqrt{a^2-b^2}=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}.$

Thus, $a=6,b=4$ and $c=2\sqrt{5}.$

(i) Length of the major axis = $2a=(2\times 6)$ units = 12 units.

Length of the minor axis = $2b=(2\times 4)$ units = 8 units.

(ii) Coordinates of the vertices are $A(-a,0)$ and $B(a,0),$ i.e., $A(-6,0)$ and $B(6,0).$

(iii) Coordinates of the foci are $F_1(-c,0)$ and $F_2(c,0),$ i.e., $F_1(-2\sqrt{5,},0)$ and $F_2(2\sqrt{5},0).$

(iv) Eccentricity, $e=\frac{c}{a}=\frac{2\sqrt{5}}{6}=\frac{\sqrt{5}}{3}.$

(v) Length of the latus rectum = $\frac{2b^2}{a}=\frac{(2\times 16)}{6}$ units = $\frac{16}{3}$ units.