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$

Answer 1

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.

http://clay6.com/mpaimg/7_toolbox10.jpg

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 : $(0,\pm a)$

$\Rightarrow (0,\pm 3)$