Question

$\text{Write the length of the latus-rectum of the hyperbola}16x^2-9y^2=144$

Answer 1

$\text{The given equation of the hyperbola is}$

$16x^2-9y^2=144$

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

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

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

$\Rightarrow \frac{x^2}{3^2}-\frac{y^2}{4^2}=1$

$Thisequationisoftheform\frac{x^{2}3}{a^2}-\frac{y^2}{b^2}=1,$

$wherea=3andb=4$

$\text{Now,the length of the latus-rectum}=\frac{2b^2}{a}$

$=2\times \frac{16}{3}$

$=\frac{32}{3}$