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 $y^2-16x^2=16.$

Answer 1

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

Clearly, the given equation represents a vertical hyperbola.

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

we get $(a^2=16and{b}^2=1)\Rightarrow (a=4andb=1).$

$\therefore c=\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17}.$

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

Length of the conjugate axis = $2b=(2\times 1)$ units = 2 units.

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

(iii) The coordinates of the foci are $F_1(0,-c)$ and $F_2(0,c),$ i.e., $F_1(0,-\sqrt{17)}$ and $F_2(0,\sqrt{17).}$

(iv) Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{17}}{4}.$

(v) Length of the latus rectum = $\frac{2b^2}{a}=\left(\frac{2\times 1}{4}\right)$ unit = $\frac{1}{2}$ unit.