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# Find the eccentricity, centre , foci , and vertices of the following hyperbolas and draw their diagrams: $\large\frac{y^{2}}{9}-\frac{x^{2}}{25}=$$1 ## 1 Answer Comment A) Toolbox: • Standard forms of equation of the hyperbola with transverse axis 2a,conjugate axis 2t with the negative sign associated with b and e=\sqrt{1+\large\frac{b^2}{a^2}},b=a\sqrt{e^2-1} • \large\frac{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
• http://clay6.com/mpaimg/5_toolbox10.jpg
• Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
• Transverse axis $y$-axis ($x=0$)
• Conjugate axis $x$-axis ($y=0$)
• End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
• Length of LR :$\large\frac{2b^2}{a}$
• Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$\large\frac{y^2}{9}-\large\frac{x^2}{25}$$=1 a^2=9,b^2=25 a=4,b=5 and x-axis is the transverse axis. Step 2: Eccentricity e=\sqrt{1+\large\frac{b^2}{a^2}} \Rightarrow \sqrt{1+\large\frac{25}{9}}=\large\frac{\sqrt{34}}{3} Step 3: Centre : Origin O(0,0) Step 4: Foci : (0,\pm ae) ae=3\times \large\frac{\sqrt{34}}{3}=$$\sqrt{34}$
$\Rightarrow (0,\pm\sqrt{34})$
Vertices : $(0,\pm a)$
$\Rightarrow (0,\pm 3)$