Question

Find the lengths of transverse axis and conjugate axis, eccentricity, the co-ordinates of foci, vertices, length of the latus-rectum, and equations of the directrices of the following hyperbola $16x^2-9y^2=-144$.

Answer 1

Step $1$: Compute the lengths of the transverse axis and conjugate axis:

The given hyperbola can be written as: $\frac{x^2}{9}-\frac{y^2}{16}=-1$.

Comparing with standard form: $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$.

$$\begin{gathered} a^2=9 \\ \Rightarrow a=3 \end{gathered}$$ and,

$$\begin{gathered} b^2=16 \\ \Rightarrow b=4 \end{gathered}$$

So we can write that,

Length of transverse axis is $2b=2\left(4\right)=8$

So we can write that,

Length of conjugate axis is $2a=2\left(3\right)=6$

Step $2$: Compute the eccentricity.

We know that eccentricity $e=\sqrt{\left(1+\frac{a^2}{b^2}\right)}$

$$\begin{gathered} \Rightarrow e=\sqrt{\left(1+\frac{3^2}{4^2}\right)} \\ =\sqrt{\frac{25}{16}} \\ =\frac{5}{4} \end{gathered}$$

Step $3$: Compute Foci and the vertices of the hyperbola.

The coordinates of the foci are:

$\left(0,\pm be\right)=\left(0,\pm 4\times \frac{5}{4}\right)=\left(0,\pm 5\right)$

The coordinates of the vertices are:

$\left(0,\pm b\right)=\left(0,\pm 4\right)$

Step $4$: Compute the length of the latus - rectum.

The length of latus - rectum is:

$L=\frac{2a^2}{b}=\frac{2{\left(3\right)}^2}{4}=\frac{9}{2}$

Step $5$: Compute the equations of directrices.

The equations of the directrices are :

$$\begin{gathered} \Rightarrow y=\pm \frac{b}{e} \\ \Rightarrow y=\pm \frac{4}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \Rightarrow y=\pm \frac{16}{5} \end{gathered}$$

Hence, the length of transverse axis is $8$, length of conjugate axis is $6$, eccentricity is $\frac{5}{4}$, foci is $\left(0,\pm 5\right)$, vertices of hyperbola are $\left(0,\pm 4\right)$, length of latus-rectum is $\frac{9}{2}$, equation of directrices is $y=\pm \frac{16}{5}$.