Question

Find the eccentricity, coordinates of the foci, equations of directrices and length of the latus-rectum of the hyperbola
(i) 9x² − 16y² = 144
(ii) 16x² − 9y² = −144
(iii) 4x² − 3y² = 36
(iv) 3x² − y² = 4
(v) 2x² − 3y² = 5.

Answer 1

(i) Equation of the hyperbola:
$9x^2-16y^2=144$
This can be rewritten in the following way:
$\frac{x^2}{16}-\frac{y^2}{9}=1$
This is the standard equation of a hyperbola, where $a^2=16\mathrm{and}{b}^2=9$.

$$\begin{gathered} \Rightarrow b^2=a^2(e^2-1) \\ \Rightarrow 9=16(e^2-1) \\ \Rightarrow e^2-1=\frac{9}{16} \\ \Rightarrow e^2=\frac{25}{16} \\ \Rightarrow e=\frac{5}{4} \end{gathered}$$
Coordinates of the foci are given by $\left(\pm ae,0\right)$, i.e. $\left(\pm 5,0\right)$.
Equation of directrices:
$x=\pm \frac{a}{e}$
$$\begin{gathered} \Rightarrow x=\pm \frac{4}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \Rightarrow 5x\pm 16=0 \end{gathered}$$
Length of the latus rectum of the hyperbola is $\frac{2b^2}{a}$.
Length of the latus rectum = $\frac{2\times 9}{4}=\frac{9}{2}$

(ii) Equation of the hyperbola:
$16x^2-9y^2=-144$
This can be rewritten in the following way:
$\frac{x^2}{9}-\frac{y^2}{16}=-1$
This is the standard equation of a hyperbola, where $a^2=9\mathrm{and}{b}^2=16$.

$$\begin{gathered} \Rightarrow a^2=b^2(e^2-1) \\ \Rightarrow 9=16(e^2-1) \\ \Rightarrow e^2-1=\frac{9}{16} \\ \Rightarrow e^2=\frac{25}{16} \\ \Rightarrow e=\frac{5}{4} \end{gathered}$$
Coordinates of foci are given by $\left(0,\pm ae\right)$, i.e. $\left(0,\pm 5\right)$.
Equation of the directrices:
$y=\pm \frac{a}{e}$
$$\begin{gathered} \Rightarrow y=\pm \frac{4}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \Rightarrow 5y\pm 16=0 \end{gathered}$$
Length of the latus rectum of the hyperbola = $\frac{2a^2}{b}$
Length of the latus rectum = $\frac{2\times 9}{4}=\frac{9}{2}$

(iii) Equation of the hyperbola:
$4x^2-3y^2=36$
This can be rewritten in the following way:
$$\begin{gathered} \frac{4x^2}{36}-\frac{3y^2}{36}=1 \\ \frac{x^2}{9}-\frac{y^2}{12}=1 \end{gathered}$$
This is the standard equation of a hyperbola, where $a^2=9\mathrm{and}{b}^2=12$.

$$\begin{gathered} \Rightarrow b^2=a^2(e^2-1) \\ \Rightarrow 12=9(e^2-1) \\ \Rightarrow e^2-1=\frac{4}{3} \\ \Rightarrow e^2=\frac{7}{3} \\ \Rightarrow e=\sqrt{\frac{7}{3}} \end{gathered}$$
Coordinates of the foci are given by $\left(\pm ae,0\right)$, i.e. $\left(\pm \sqrt{21},0\right)$.
Equation of the directrices:
$x=\pm \frac{a}{e}$
$$\begin{gathered} \Rightarrow x=\pm \frac{3}{{\displaystyle \sqrt{\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}} \\ \Rightarrow \sqrt{7}x\pm 3\sqrt{3}=0 \end{gathered}$$
Length of the latus rectum of the hyperbola is $\frac{2b^2}{a}$.
$\Rightarrow \frac{2\times 12}{3}=8$

(iv) Equation of the hyperbola:
$3x^2-y^2=4$
This can be rewritten in the following way:
$$\begin{gathered} \frac{3x^2}{4}-\frac{y^2}{4}=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-\frac{y^2}{4}=1 \end{gathered}$$
This is the standard equation of a hyperbola, where $a^2=\frac{4}{3}\mathrm{and}{b}^2=4$.

$$\begin{gathered} \Rightarrow b^2=a^2(e^2-1) \\ \Rightarrow 4=\frac{4}{3}(e^2-1) \\ \Rightarrow e^2-1=3 \\ \Rightarrow e^2=4 \\ \Rightarrow e=2 \end{gathered}$$
Coordinates of the foci are given by $\left(\pm ae,0\right)$, i.e. $\left(\pm \frac{4\sqrt{3}}{3},0\right)$.
Equation of the directrices:
$x=\pm \frac{a}{e}$
$$\begin{gathered} x=\pm \frac{\sqrt{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{2} \\ \Rightarrow \sqrt{3}x\pm 1=0 \end{gathered}$$
Length of the latus rectum of the hyperbola = $\frac{2b^2}{a}$
$\Rightarrow \frac{2\times 4}{\sqrt{{\displaystyle \raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}=4\sqrt{3}$

(v) Equation of the hyperbola:
$2x^2-3y^2=5$
This can be rewritten in the following manner:
$$\begin{gathered} \frac{2x^2}{5}-\frac{3y^2}{5}=1 \\ \Rightarrow \frac{x^2}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-\frac{y^2}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=1 \end{gathered}$$
This is the standard equation of a hyperbola, where $a^2=\frac{5}{2}\mathrm{and}{b}^2=\frac{5}{3}$.

$$\begin{gathered} \Rightarrow b^2=a^2(e^2-1) \\ \Rightarrow \frac{5}{3}=\frac{5}{2}(e^2-1) \\ \Rightarrow e^2-1=\frac{2}{3} \\ \Rightarrow e^2=\frac{5}{3} \\ \Rightarrow e=\sqrt{\frac{5}{3}} \end{gathered}$$
Coordinates of the foci are given by $\left(\pm ae,0\right)$, i.e. $\left(\pm \frac{5\sqrt{6}}{6},0\right)$.
Equation of the directrices:
$x=\pm \frac{a}{e}$
$$\begin{gathered} x=\pm \frac{\sqrt{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \\ \Rightarrow x=\pm \frac{\sqrt{3}}{\sqrt{2}} \\ \Rightarrow \sqrt{2}x\pm \sqrt{3}=0 \end{gathered}$$
Length of the latus rectum of the hyperbola is $\frac{2b^2}{a}$.

$\Rightarrow \frac{2\times \left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}{\sqrt{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}=\frac{10}{3}\sqrt{\frac{2}{5}}$