Question

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

Answer 1

The equation $16x^2-9y^2=144.$ can be written as $\frac{x^2}{9}-\frac{y^2}{16}=-1$

This is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\therefore a^2=9,b^2=16\Rightarrow a=3,b=4$

Length of transverse axis : The length of transverse axis = 2b = 8

Length of conjugate axis : The length of conjugate axis = 2a =6

Eccentricity : $\sqrt{(1+\frac{a^2}{b^2})}=\sqrt{(1+\frac{9}{16})}=\frac{5}{4}$

Foci : The co-ordinates of the foci are $(0,\pm be)i.e.,(0,\pm 5)$

Vertices : The co-ordinates of the vertices are $(0,\pm b)i.e.,(0,\pm 4)$

Length of latus-rectum : The length of latus-rectum $=\frac{2a^2}{b}=\frac{2{\left(3\right)}^2}{4}=\frac{9}{2}$

Equation of directrices : The equation of directrices are $y=\pm \frac{b}{e}\Rightarrow y=\pm \frac{4}{\left(5/4\right)}\Rightarrow y=\pm \frac{16}{5}$