Question

Find the lengths of the axes; the coordinates of the vertices and the foci; the eccentricity and length of the latus rectum of the hyperbola $9x^2-16y^2=144.$

Answer 1

$9x^2-16y^2=144\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1.$

Thus, the euqation of the given hyperbola is $\frac{x^2}{16}-\frac{y^2}{9}=1.$

Comparing the given equation with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$

we get $a^2=16$ and $b^2=9.$

$\therefore a=4,b=3$ and $c=\sqrt{a^2+b^2}=\sqrt{16+9}=\sqrt{25}=5.$

(i) Length of the transverse axis = $2a=(2\times 4)$ units = 8 units.

(ii) The coordinates of the vertices are $A(-a,0)$ and $B(a,0),$ i.e., $A(-4,0)$ and $B(4,0).$

(iii) The coordinates of the foci are $F_1(-c,0)$ and $F_2(c,0),$ i.e., $F_1(-5,0)$ and $F_2(5,0).$

(iv) Ecentricity, $e=\frac{c}{a}=\frac{5}{4}.$

(v) Length of the latus rectum = $\frac{2b^2}{a}=\left(\frac{2\times 9}{4}\right)$ units = $\frac{9}{2}$ units.