Question

Find the equation of the hyperbola with centre at the origin, length of the transverse axis 6 and one focus at (0, 4).

Answer 1

Since the coordinates of the foci are of the form $(0,\pm c),$ it is a case of vertical hyperbola.

Let its equation be $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1.$

Clearly, $c=4[\because (0,c)=(0,4\left)\right]$

Length of the transverse axis = $6\iff 2a=6\iff a=3.$

Also, $c^2=(a^2+b^2)\iff b^2-(c^2-a^2)=(4^2-3^2)=(16-9)=7.$

Thus, $a^2=3^2$ and $b^2=7.$

Hence, the required equation is $\frac{y^2}{9}-\frac{x^2}{7}=1.$