Question

$\text{Find the ecentrictity, coordinates of foci, equations of directrices and length of the latus-rectum of the hyperbola}\left(i\right)9x^2-16y^2=144\left(ii\right)16x^2-9y^2=-144\left(iiii\right)4x^2-3y^2=36\left(iv\right)3x^2-y^2=4\left(v\right)2x^2-3y^2=5$

Answer 1

$\left(i\right)Wehave,9x^2-16y^2=144\Rightarrow \frac{9x^2}{144}-\frac{16y^2}{144}=1\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1\text{This is of the form}\Rightarrow \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,Where{a}^2=16and{b}^2=9\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\text{Foci : The coordinates of the foci are}(\pm ae,0)ie.,(\pm 5,0)\text{Equations of the directrices : The equations of the directices are}x=\pm \frac{a}{e}ie.,x=\pm \frac{16}{5}\because 5x=\pm 16\Rightarrow 5x\mp 16=0\text{Length of latus-rectum : The length of the latus-rectum}=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}\left(ii\right)Wehave,16x^2-9y^2=-144\Rightarrow \frac{16x^2}{144}-\frac{9y^2}{144}=-1\Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=-1\text{This is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,where{a}^2=9and{b}^2=16\therefore a=3andb=4\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{5}{4}\text{Foci : The coordinates of the foci are}(0,\pm be)\therefore (0,\pm be)=(0,\pm 4\times \frac{5}{4})=(0,\pm 5)\therefore \text{the coordinates of the foci are}(0,\pm 5)\text{Equations of the directices : The equations of the directrices are}y=\frac{\pm b}{e}\Rightarrow y=\pm \frac{4}{5}=\pm \frac{16}{5}\Rightarrow 5y\mp 16=0\text{Latus-rectum : The length of the latus-rectum}=\frac{2a^2}{b}=\frac{2\times 9}{4}=\frac{9}{2}\left(iiii\right)Wehave,4x^2-3y^2=36\Rightarrow \frac{4x^2}{36}-\frac{3y^2}{36}=1\Rightarrow \frac{x^2}{9}-\frac{y^2}{12}=1\text{This is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,where{a}^2=9and{b}^2=12\therefore a=3andb=\sqrt{12}=2\sqrt{3}\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{12}{9}}=\sqrt{1+\frac{4}{3}}=\sqrt{\frac{7}{3}}\text{Foci : The coordinates of the foci are}(0,\pm ae,0)\text{The equations of the directrices are}\sqrt{7}x\pm 3\sqrt{3}=0\text{Latus-rectum : The length of the latus-rectum}=\frac{2b^2}{a}=\frac{1\times 12}{3}=8\pm ac=\pm 3\times \frac{\sqrt{7}}{\sqrt{3}}=\pm 3\times \frac{\sqrt{7}}{\sqrt{3}}=\pm \sqrt{3}\times \sqrt{7}=\pm \sqrt{21}(\pm ac,0)=(\pm \sqrt{21,0})\therefore \text{The coordinates of the foci are}(\pm \sqrt{21,0})\text{Equation of the directrices : The equation of the directorices are}x=\frac{\pm a}{e}x=\pm 3\times \frac{1}{\frac{\sqrt{7}}{\sqrt{3}}}=\frac{\pm 3\sqrt{3}}{\sqrt{7}}\Rightarrow \sqrt{7}x\pm 3\sqrt{3}=0\text{The equations of the directrices are}\sqrt{7}x\pm 3\sqrt{3}=0\left(iv\right)Wehave3x^2-y^2=4\Rightarrow \frac{3x^2}{4}-\frac{y^2}{4}=1\frac{x^2}{\frac{4}{3}}-\frac{y^2}{4}=1\Rightarrow \frac{x^2}{\left(\frac{2}{\sqrt{3}}\right)}-\frac{y^2}{2^2}=1\text{This is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,wherea=\frac{2}{\sqrt{3}}andb=2\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{\frac{4}{3}}}=\sqrt{1+3}=\sqrt{4}=2\text{Foci : The coordinates of the foci are}(\pm ae,0)\therefore \pm ae=\pm \frac{2}{\sqrt{3}}\times 2=\pm \frac{4}{\sqrt{3}}\text{The coordinates of the foci are}(\pm \frac{4}{\sqrt{3}},0)\text{Equations of the directrices : The equations of the directrices are}x=\pm \frac{a}{e}=\pm \frac{2}{\frac{\sqrt{3}}{2}}=\pm \frac{1}{\sqrt{3}}\Rightarrow \sqrt{3}x\pm 1=0\text{Latus-rectum : The length of the latus-rectum}=\frac{2b^2}{a}\therefore \frac{2b^2}{a}=2\times \frac{4}{2}=4\sqrt{3}.\left(v\right)Wehave2x^2-3y^2=5\Rightarrow \frac{2x^2}{5}-\frac{3y^2}{5}=1\frac{x^2}{\frac{5}{2}}-\frac{y^2}{\frac{5}{3}}=1\Rightarrow \frac{x^2}{{\left(\frac{5}{\sqrt{2}}\right)}^2}-\frac{y^2}{\left(\sqrt{\frac{5}{3}}\right)}=1\text{This is of the form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,wherea=\sqrt{\frac{5}{2}}andb=\sqrt{\frac{5}{3}}\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{\frac{5}{3}}{\frac{5}{2}}}=\sqrt{1+\frac{5}{3}\times \frac{2}{5}}=\sqrt{1+\frac{2}{3}}=\sqrt{\frac{5}{3}}\text{Foci : The coordinates of the foci are}(\pm ae,0)\therefore \pm ae=\pm \sqrt{\frac{5}{2}}\times \sqrt{\frac{5}{3}}=\pm \frac{5}{\sqrt{6}}\therefore Foci:(\pm \frac{5}{\sqrt{6}},0)\text{Equations of the directirices : The equations of the directireices are}x=\pm \frac{a}{e}=\pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}}=\pm \frac{\sqrt{5}}{\sqrt{2}}\times \frac{\sqrt{3}}{\sqrt{5}}=\pm \sqrt{\frac{3}{2}}\Rightarrow x=\pm \frac{\sqrt{3}}{\sqrt{2}}\Rightarrow \sqrt{2}x\pm \sqrt{3}=0\text{Latus-rectum : The length of the latus-rectum}=\frac{2b^2}{a}\therefore \frac{2b^2}{a}=2\times \frac{3}{\sqrt{\frac{5}{2}}}=\frac{10\times \sqrt{2}}{3\times \sqrt{5}}=\frac{10}{3}\sqrt{\frac{2}{5}}$