NIMCET 2017 Question Paper with solutions

We have,
Focus = (0, 4) and length of transverse axis = 6
∴ be = 4 and 26 = 6
⇒ be = 4 and b = 3
⇒ e = $\frac{4}{3}$ and b = 3
also ${\mathit{e}}^2$ = $1+\frac{{\mathit{a}}^2}{{\mathit{b}}^2}$
$\frac{16}{9}$ = 1 + $\frac{{\mathit{a}}^2}{9}$
⇒ ${\mathit{a}}^2$ = 7
∴ Equation of hyperbola is
$\frac{{\mathit{x}}^2}{{\mathit{a}}^2}-\frac{{\mathit{y}}^2}{{\mathit{b}}^2}$ = - 1
⇒ $\frac{{\mathit{x}}^2}{7}-\frac{{\mathit{y}}^2}{9}$ = - 1
⇒ $\frac{{\mathit{y}}^2}{9}-\frac{{\mathit{x}}^2}{7}$ = 1