Question

Find the centre, focii, eccentricity and length of the latus rectum of the hyperbola $16y^2-9x^2=144$

Answer 1

$Thegivenequationis$ $\frac{x^2}{9}-\frac{y^2}{16}=1$

$Theaboveequationisoftheform$ $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1thus$

$\begin{array}{c}{a}^2=9\\ a=3\\ 8b^2=16\\ b=4\end{array}$

$\begin{array}{cc}\text{Also,}& \\ c^2=a^2+b^2& \\ c^2=9+16& \\ c^2=25& \\ c=5& \\ \text{So, Co-ordinate of foci}=(\pm c,0)& \\ \text{Thus, Co-ordinate of foci are}(5,0)\&(-5,0)& \end{array}$

$\begin{array}{cc}\text{Vertices}& =(\pm a,0)\\ & =(\pm 3,0)\\ \text{Thus, Vertices are}(3,0)\&(-3,0)& \end{array}$

$\begin{array}{cc}\text{Eccentricity}e& =\frac{c}{a}\\ & =\frac{5}{3}\\ \text{The latus rectum}& =\frac{2b^2}{a}\\ & =\frac{2\times 16}{3}\\ & =\frac{32}{3}\end{array}$