Question 12 Marks

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola
$\frac{x^2}{4}-\frac{y^2}{12}=1$
$\frac{y^2}{9}-\frac{x^2}{16}=1$

Answer

The equation of the hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$.
Comparing this with equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
We have $a^2=4, b^2=12$
$
\therefore a=2, b=2 \sqrt{3}
$
Length of transverse axis $=2 a=2(2)=4$
Length of conjugate axis $=2 b=2(2 \sqrt{3})$ $=4 \sqrt{3}$
Eccentricity $b^2=a^2\left(e^2-1\right)$
$
\begin{aligned}
& \therefore \mathrm{e}=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{4+12}}{2}=\frac{\sqrt{16}}{2}=\frac{4}{2}=2 \\
& \quad(\because \mathrm{e}>0) \\
& a e=2(2)=4 \\
& \therefore \text { foci }( \pm a e, 0) \text { are }( \pm 4,0) \\
& \frac{a}{e}=\frac{2}{2}=1
\end{aligned}
$
$\therefore$ the equations of directrices $x= \pm \frac{a}{e}$ are $x= \pm 1$.
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2(12)}{2}=12$
(ii) The equation of the hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$ Comparing this with the equation
$
\frac{y^2}{b^2}-\frac{x^2}{a^2}=1
$
We have $a^2=16, b^2=9$
$
\therefore a=4, b=3
$
Length of transverse axis $=2 b=2(3)=6$
Length of conjugate axis $=2 a=2(4)=8$
Eccentricity $a^2=b^2\left(e^2-1\right)$
$
\begin{aligned}
& \therefore \mathrm{e}=\frac{\sqrt{a^2+b^2}}{b}=\frac{\sqrt{16+9}}{3}=\frac{\sqrt{25}}{3}=\frac{5}{3} \\
& (\because \mathrm{e}>0) \\
& b e=3\left(\frac{5}{3}\right)=5 \\
& \therefore \text { focii }(0, \pm b e) \text { are }(0,+5) \\
& \frac{b}{e}=\frac{3}{5 / 3}=\frac{9}{5} \\
&
\end{aligned}
$
$\therefore$ the equations of directrices $y= \pm \frac{b}{e}$ are $y= \pm \frac{9}{5}$
Length of latus rectum $=\frac{2 a^2}{b}=\frac{2(16)}{3}=\frac{32}{3}$

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