Class 11: Hyperbola – Exercise 27.1

$\displaystyle \textbf{Question 1: }\text{The equation of the directrix of a hyperbola is }x-y+3=0.\text{ Its focus is}$
$\displaystyle (-1,1)\text{ and its eccentricity is }3.\text{ Find the equation of the hyperbola.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }S(-1,1)\text{ be the focus and }P(x,y)\text{ be any point on the hyperbola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix }x-y+3=0.$
$\displaystyle \text{By the focus-directrix definition of a hyperbola,}$
$\displaystyle SP=e\cdot PM$
$\displaystyle \therefore SP^2=e^2PM^2$
$\displaystyle (x+1)^2+(y-1)^2=3^2\left(\frac{x-y+3}{\sqrt{1^2+(-1)^2}}\right)^2$
$\displaystyle (x+1)^2+(y-1)^2=\frac{9}{2}(x-y+3)^2$
$\displaystyle 2\left[x^2+y^2+2x-2y+2\right]=9(x-y+3)^2$
$\displaystyle 2x^2+2y^2+4x-4y+4$
$\displaystyle =9\left(x^2+y^2-2xy+6x-6y+9\right)$
$\displaystyle 2x^2+2y^2+4x-4y+4$
$\displaystyle =9x^2+9y^2-18xy+54x-54y+81$
$\displaystyle \therefore 7x^2+7y^2-18xy+50x-50y+77=0$
$\displaystyle \therefore \text{The required equation of the hyperbola is}$
$\displaystyle 7x^2+7y^2-18xy+50x-50y+77=0.$

$\displaystyle \textbf{Question 2: }\text{Find the equation of the hyperbola whose}$
$\displaystyle \text{(i) focus is }(0,3),\text{ directrix is }x+y-1=0\text{ and eccentricity is }2;$
$\displaystyle \text{(ii) focus is }(1,1),\text{ directrix is }3x+4y+8=0\text{ and eccentricity is }2;$
$\displaystyle \text{(iii) focus is }(1,1),\text{ directrix is }2x+y=1\text{ and eccentricity is }\sqrt{3};$
$\displaystyle \text{(iv) focus is }(2,-1),\text{ directrix is }2x+3y=1\text{ and eccentricity is }2;$
$\displaystyle \text{(v) focus is }(a,0),\text{ directrix is }2x-y+a=0\text{ and eccentricity is }\frac{4}{3};$
$\displaystyle \text{(vi) focus is }(2,2),\text{ directrix is }x+y=9\text{ and eccentricity is }2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }S\text{ be the given focus and }P(x,y)\text{ be any point on the hyperbola.}$
$\displaystyle \text{If }PM\text{ is perpendicular to the directrix, then }SP=e\cdot PM.$
$\displaystyle \text{Therefore, }SP^2=e^2PM^2.$
$\displaystyle \text{(i) Focus }S(0,3),\text{ directrix }x+y-1=0\text{ and }e=2.$
$\displaystyle x^2+(y-3)^2=2^2\left(\frac{x+y-1}{\sqrt{1^2+1^2}}\right)^2$
$\displaystyle x^2+y^2-6y+9=2(x+y-1)^2$
$\displaystyle x^2+y^2-6y+9=2(x^2+y^2+2xy-2x-2y+1)$
$\displaystyle x^2+y^2-6y+9=2x^2+2y^2+4xy-4x-4y+2$
$\displaystyle \therefore x^2+y^2+4xy-4x+2y-7=0.$
$\displaystyle \text{Hence, the required equation is }x^2+y^2+4xy-4x+2y-7=0.$

$\displaystyle \text{(ii) Focus }S(1,1),\text{ directrix }3x+4y+8=0\text{ and }e=2.$
$\displaystyle (x-1)^2+(y-1)^2=2^2\left(\frac{3x+4y+8}{\sqrt{3^2+4^2}}\right)^2$
$\displaystyle x^2+y^2-2x-2y+2=\frac{4}{25}(3x+4y+8)^2$
$\displaystyle 25(x^2+y^2-2x-2y+2)=4(3x+4y+8)^2$
$\displaystyle 25x^2+25y^2-50x-50y+50$
$\displaystyle =4(9x^2+16y^2+24xy+48x+64y+64)$
$\displaystyle 25x^2+25y^2-50x-50y+50$
$\displaystyle =36x^2+64y^2+96xy+192x+256y+256$
$\displaystyle \therefore 11x^2+39y^2+96xy+242x+306y+206=0.$
$\displaystyle \text{Hence, the required equation is}$
$\displaystyle 11x^2+39y^2+96xy+242x+306y+206=0.$

$\displaystyle \text{(iii) Focus }S(1,1),\text{ directrix }2x+y-1=0\text{ and }e=\sqrt{3}.$
$\displaystyle (x-1)^2+(y-1)^2=(\sqrt{3})^2\left(\frac{2x+y-1}{\sqrt{2^2+1^2}}\right)^2$
$\displaystyle x^2+y^2-2x-2y+2=\frac{3}{5}(2x+y-1)^2$
$\displaystyle 5(x^2+y^2-2x-2y+2)=3(2x+y-1)^2$
$\displaystyle 5x^2+5y^2-10x-10y+10$
$\displaystyle =3(4x^2+y^2+4xy-4x-2y+1)$
$\displaystyle 5x^2+5y^2-10x-10y+10$
$\displaystyle =12x^2+3y^2+12xy-12x-6y+3$
$\displaystyle \therefore 7x^2-2y^2+12xy-2x+4y-7=0.$
$\displaystyle \text{Hence, the required equation is}$
$\displaystyle 7x^2-2y^2+12xy-2x+4y-7=0.$

$\displaystyle \text{(iv) Focus }S(2,-1),\text{ directrix }2x+3y-1=0\text{ and }e=2.$
$\displaystyle (x-2)^2+(y+1)^2=2^2\left(\frac{2x+3y-1}{\sqrt{2^2+3^2}}\right)^2$
$\displaystyle x^2+y^2-4x+2y+5=\frac{4}{13}(2x+3y-1)^2$
$\displaystyle 13(x^2+y^2-4x+2y+5)=4(2x+3y-1)^2$
$\displaystyle 13x^2+13y^2-52x+26y+65$
$\displaystyle =4(4x^2+9y^2+12xy-4x-6y+1)$
$\displaystyle 13x^2+13y^2-52x+26y+65$
$\displaystyle =16x^2+36y^2+48xy-16x-24y+4$
$\displaystyle \therefore 3x^2+23y^2+48xy+36x-50y-61=0.$
$\displaystyle \text{Hence, the required equation is}$
$\displaystyle 3x^2+23y^2+48xy+36x-50y-61=0.$

$\displaystyle \text{(v) Focus }S(a,0),\text{ directrix }2x-y+a=0\text{ and }e=\frac{4}{3}.$
$\displaystyle (x-a)^2+y^2=\left(\frac{4}{3}\right)^2\left(\frac{2x-y+a}{\sqrt{2^2+(-1)^2}}\right)^2$
$\displaystyle x^2+y^2-2ax+a^2=\frac{16}{45}(2x-y+a)^2$
$\displaystyle 45(x^2+y^2-2ax+a^2)=16(2x-y+a)^2$
$\displaystyle 45x^2+45y^2-90ax+45a^2$
$\displaystyle =16(4x^2+y^2+a^2-4xy+4ax-2ay)$
$\displaystyle 45x^2+45y^2-90ax+45a^2$
$\displaystyle =64x^2+16y^2+16a^2-64xy+64ax-32ay$
$\displaystyle \therefore 19x^2-29y^2-64xy+154ax-32ay-29a^2=0.$
$\displaystyle \text{Hence, the required equation is}$
$\displaystyle 19x^2-29y^2-64xy+154ax-32ay-29a^2=0.$

$\displaystyle \text{(vi) Focus }S(2,2),\text{ directrix }x+y-9=0\text{ and }e=2.$
$\displaystyle (x-2)^2+(y-2)^2=2^2\left(\frac{x+y-9}{\sqrt{1^2+1^2}}\right)^2$
$\displaystyle x^2+y^2-4x-4y+8=2(x+y-9)^2$
$\displaystyle x^2+y^2-4x-4y+8$
$\displaystyle =2(x^2+y^2+2xy-18x-18y+81)$
$\displaystyle x^2+y^2-4x-4y+8$
$\displaystyle =2x^2+2y^2+4xy-36x-36y+162$
$\displaystyle \therefore x^2+y^2+4xy-32x-32y+154=0.$
$\displaystyle \text{Hence, the required equation is}$
$\displaystyle x^2+y^2+4xy-32x-32y+154=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the eccentricity, coordinates of the foci, equations of the directrices and}$
$\displaystyle \text{length of the latus rectum of each of the following hyperbolas:}$
$\displaystyle \text{(i) }9x^2-16y^2=144\qquad \text{(ii) }16x^2-9y^2=-144$
$\displaystyle \text{(iii) }4x^2-3y^2=36\qquad \text{(iv) }3x^2-y^2=4$
$\displaystyle \text{(v) }2x^2-3y^2=5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }9x^2-16y^2=144$
$\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ we get }a^2=16,\ b^2=9.$
$\displaystyle \therefore a=4,\qquad b=3$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
$\displaystyle \text{Foci: }(\pm ae,0)=\left(\pm4\times\frac{5}{4},0\right)=(\pm5,0)$
$\displaystyle \text{Directrices: }x=\pm\frac{a}{e}=\pm\frac{4}{5/4}=\pm\frac{16}{5}$
$\displaystyle \text{Length of latus rectum: }\frac{2b^2}{a}=\frac{2\times9}{4}=\frac{9}{2}$
$\displaystyle \\$

$\displaystyle \text{(ii) Given, }16x^2-9y^2=-144$
$\displaystyle \frac{y^2}{16}-\frac{x^2}{9}=1$
$\displaystyle \text{Comparing with }\frac{y^2}{a^2}-\frac{x^2}{b^2}=1,\text{ we get }a^2=16,\ b^2=9.$
$\displaystyle \therefore a=4,\qquad b=3$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
$\displaystyle \text{Foci: }(0,\pm ae)=\left(0,\pm4\times\frac{5}{4}\right)=(0,\pm5)$
$\displaystyle \text{Directrices: }y=\pm\frac{a}{e}=\pm\frac{4}{5/4}=\pm\frac{16}{5}$
$\displaystyle \text{Length of latus rectum: }\frac{2b^2}{a}=\frac{2\times9}{4}=\frac{9}{2}$
$\displaystyle \\$

$\displaystyle \text{(iii) Given, }4x^2-3y^2=36$
$\displaystyle \frac{x^2}{9}-\frac{y^2}{12}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ we get }a^2=9,\ b^2=12.$
$\displaystyle \therefore a=3,\qquad b=2\sqrt{3}$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{12}{9}}=\sqrt{\frac{7}{3}}$
$\displaystyle \text{Foci: }(\pm ae,0)=\left(\pm3\sqrt{\frac{7}{3}},0\right)=(\pm\sqrt{21},0)$
$\displaystyle \text{Directrices: }x=\pm\frac{a}{e}=\pm\frac{3}{\sqrt{7/3}}$
$\displaystyle \therefore x=\pm\frac{3\sqrt{3}}{\sqrt{7}}$
$\displaystyle \text{Length of latus rectum: }\frac{2b^2}{a}=\frac{2\times12}{3}=8$
$\displaystyle \\$

$\displaystyle \text{(iv) Given, }3x^2-y^2=4$
$\displaystyle \frac{x^2}{4/3}-\frac{y^2}{4}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ we get }a^2=\frac{4}{3},\ b^2=4.$
$\displaystyle \therefore a=\frac{2}{\sqrt{3}},\qquad b=2$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{4}{4/3}}=2$
$\displaystyle \text{Foci: }(\pm ae,0)=\left(\pm\frac{2}{\sqrt{3}}\times2,0\right)$
$\displaystyle =\left(\pm\frac{4}{\sqrt{3}},0\right)$
$\displaystyle \text{Directrices: }x=\pm\frac{a}{e}=\pm\frac{2/\sqrt{3}}{2}=\pm\frac{1}{\sqrt{3}}$
$\displaystyle \text{Length of latus rectum: }\frac{2b^2}{a}=\frac{2\times4}{2/\sqrt{3}}=4\sqrt{3}$
$\displaystyle \\$

$\displaystyle \text{(v) Given, }2x^2-3y^2=5$
$\displaystyle \frac{x^2}{5/2}-\frac{y^2}{5/3}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ we get }a^2=\frac{5}{2},\ b^2=\frac{5}{3}.$
$\displaystyle \therefore a=\sqrt{\frac{5}{2}},\qquad b=\sqrt{\frac{5}{3}}$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{5/3}{5/2}}=\sqrt{\frac{5}{3}}$
$\displaystyle \text{Foci: }(\pm ae,0)=\left(\pm\sqrt{\frac{5}{2}}\sqrt{\frac{5}{3}},0\right)$
$\displaystyle =\left(\pm\frac{5}{\sqrt{6}},0\right)$
$\displaystyle \text{Directrices: }x=\pm\frac{a}{e}=\pm\frac{\sqrt{5/2}}{\sqrt{5/3}}$
$\displaystyle \therefore x=\pm\sqrt{\frac{3}{2}}$
$\displaystyle \text{Length of latus rectum: }\frac{2b^2}{a}$
$\displaystyle =\frac{2(5/3)}{\sqrt{5/2}}=\frac{2\sqrt{10}}{3}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the axes, eccentricity, latus rectum and coordinates of the foci of the}$
$\displaystyle \text{hyperbola }25x^2-36y^2=225.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }25x^2-36y^2=225$
$\displaystyle \frac{25x^2}{225}-\frac{36y^2}{225}=1$
$\displaystyle \frac{x^2}{9}-\frac{y^2}{25/4}=1$
$\displaystyle \text{Comparing with }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\text{ we get }a^2=9,\ b^2=\frac{25}{4}.$
$\displaystyle \therefore a=3,\qquad b=\frac{5}{2}$
$\displaystyle \text{Transverse axis: The transverse axis is the }x\text{-axis.}$
$\displaystyle \text{Length of the transverse axis: }2a=2\times3=6$
$\displaystyle \text{Conjugate axis: The conjugate axis is the }y\text{-axis.}$
$\displaystyle \text{Length of the conjugate axis: }2b=2\times\frac{5}{2}=5$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{25/4}{9}}=\sqrt{\frac{61}{36}}=\frac{\sqrt{61}}{6}$
$\displaystyle \text{Length of the latus rectum: }\frac{2b^2}{a}$
$\displaystyle =\frac{2(25/4)}{3}=\frac{25}{6}$
$\displaystyle \text{Foci: }(\pm ae,0)=\left(\pm3\times\frac{\sqrt{61}}{6},0\right)$
$\displaystyle =\left(\pm\frac{\sqrt{61}}{2},0\right)$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find the centre, eccentricity, foci and directrices of each of the following}$
$\displaystyle \text{hyperbolas:}$
$\displaystyle \text{(i) }16x^2-9y^2+32x+36y-164=0$
$\displaystyle \text{(ii) }x^2-y^2+4x=0\qquad\text{(iii) }x^2-3y^2-2x=8$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, }16x^2-9y^2+32x+36y-164=0$
$\displaystyle 16(x^2+2x)-9(y^2-4y)=164$
$\displaystyle 16(x^2+2x+1)-9(y^2-4y+4)=164+16-36$
$\displaystyle 16(x+1)^2-9(y-2)^2=144$
$\displaystyle \frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=-1,\qquad k=2,\qquad a^2=9,\qquad b^2=16$
$\displaystyle \therefore a=3,\qquad b=4$
$\displaystyle \text{Centre: }(-1,2)$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
$\displaystyle ae=3\times\frac{5}{3}=5$
$\displaystyle \text{Foci: }(h\pm ae,k)=(-1\pm5,2)$
$\displaystyle \therefore \text{The foci are }(4,2)\text{ and }(-6,2).$
$\displaystyle \text{Directrices: }x=h\pm\frac{a}{e}$
$\displaystyle x=-1\pm\frac{3}{5/3}=-1\pm\frac{9}{5}$
$\displaystyle \therefore \text{The directrices are }x=\frac{4}{5}\text{ and }x=-\frac{14}{5}.$

$\displaystyle \text{(ii) Given, }x^2-y^2+4x=0$
$\displaystyle x^2+4x+4-y^2=4$
$\displaystyle (x+2)^2-y^2=4$
$\displaystyle \frac{(x+2)^2}{4}-\frac{y^2}{4}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=-2,\qquad k=0,\qquad a^2=4,\qquad b^2=4$
$\displaystyle \therefore a=2,\qquad b=2$
$\displaystyle \text{Centre: }(-2,0)$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{4}{4}}=\sqrt{2}$
$\displaystyle ae=2\sqrt{2}$
$\displaystyle \text{Foci: }(h\pm ae,k)=(-2\pm2\sqrt{2},0)$
$\displaystyle \text{Directrices: }x=h\pm\frac{a}{e}$
$\displaystyle x=-2\pm\frac{2}{\sqrt{2}}=-2\pm\sqrt{2}$
$\displaystyle \therefore \text{The directrices are }x=-2\pm\sqrt{2}.$

$\displaystyle \text{(iii) Given, }x^2-3y^2-2x=8$
$\displaystyle x^2-2x+1-3y^2=9$
$\displaystyle (x-1)^2-3y^2=9$
$\displaystyle \frac{(x-1)^2}{9}-\frac{y^2}{3}=1$
$\displaystyle \text{Comparing with }\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,$
$\displaystyle h=1,\qquad k=0,\qquad a^2=9,\qquad b^2=3$
$\displaystyle \therefore a=3,\qquad b=\sqrt{3}$
$\displaystyle \text{Centre: }(1,0)$
$\displaystyle \text{Eccentricity: }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{3}{9}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$
$\displaystyle ae=3\times\frac{2}{\sqrt{3}}=2\sqrt{3}$
$\displaystyle \text{Foci: }(h\pm ae,k)=(1\pm2\sqrt{3},0)$
$\displaystyle \text{Directrices: }x=h\pm\frac{a}{e}$
$\displaystyle x=1\pm\frac{3}{2/\sqrt{3}}=1\pm\frac{3\sqrt{3}}{2}$
$\displaystyle \therefore \text{The directrices are }x=1\pm\frac{3\sqrt{3}}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the equation of the hyperbola, referred to its principal axes as axes of}$
$\displaystyle \text{coordinates, in each of the following cases:}$
$\displaystyle \text{(i) the distance between the foci is }16\text{ and the eccentricity is }\sqrt{2};$
$\displaystyle \text{(ii) the length of the conjugate axis is }5\text{ and the distance between the foci is }13;$
$\displaystyle \text{(iii) the length of the conjugate axis is }7\text{ and the hyperbola passes through }(3,-2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, the distance between the foci is }16\text{ and }e=\sqrt{2}.$
$\displaystyle \text{Distance between the foci}=2ae$
$\displaystyle 2ae=16$
$\displaystyle ae=8$
$\displaystyle a\sqrt{2}=8$
$\displaystyle a=4\sqrt{2}$
$\displaystyle \therefore a^2=32$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =32(2-1)=32$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{32}-\frac{y^2}{32}=1$
$\displaystyle \therefore x^2-y^2=32.$
$\displaystyle \\$

$\displaystyle \text{(ii) Given, the length of the conjugate axis is }5\text{ and the distance between the foci is }13.$
$\displaystyle 2b=5$
$\displaystyle b=\frac{5}{2}$
$\displaystyle \therefore b^2=\frac{25}{4}$
$\displaystyle 2ae=13$
$\displaystyle ae=\frac{13}{2}$
$\displaystyle \text{Since }a^2e^2=a^2+b^2,$
$\displaystyle \left(\frac{13}{2}\right)^2=a^2+\left(\frac{5}{2}\right)^2$
$\displaystyle \frac{169}{4}=a^2+\frac{25}{4}$
$\displaystyle a^2=\frac{144}{4}=36$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{36}-\frac{y^2}{25/4}=1$
$\displaystyle \frac{x^2}{36}-\frac{4y^2}{25}=1$
$\displaystyle \therefore 25x^2-144y^2=900.$
$\displaystyle \\$

$\displaystyle \text{(iii) Given, the length of the conjugate axis is }7.$
$\displaystyle 2b=7$
$\displaystyle b=\frac{7}{2}$
$\displaystyle \therefore b^2=\frac{49}{4}$
$\displaystyle \text{Let the equation of the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \therefore \frac{x^2}{a^2}-\frac{4y^2}{49}=1$
$\displaystyle \text{Since the hyperbola passes through }(3,-2),$
$\displaystyle \frac{3^2}{a^2}-\frac{4(-2)^2}{49}=1$
$\displaystyle \frac{9}{a^2}-\frac{16}{49}=1$
$\displaystyle \frac{9}{a^2}=\frac{65}{49}$
$\displaystyle a^2=\frac{441}{65}$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{441/65}-\frac{y^2}{49/4}=1$
$\displaystyle \frac{65x^2}{441}-\frac{4y^2}{49}=1$
$\displaystyle \therefore 65x^2-36y^2=441.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the equation of the hyperbola whose:}$
$\displaystyle \text{(i) foci are }(6,4)\text{ and }(-4,4)\text{ and eccentricity is }2;$
$\displaystyle \text{(ii) vertices are }(-8,-1)\text{ and }(16,-1)\text{ and one focus is }(17,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, the foci are }(6,4)\text{ and }(-4,4)\text{ and }e=2.$
$\displaystyle \text{The centre is the midpoint of the line joining the foci.}$
$\displaystyle \therefore (h,k)=\left(\frac{6+(-4)}{2},\frac{4+4}{2}\right)=(1,4)$
$\displaystyle \text{Since the foci have the same }y\text{-coordinate, the transverse axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{(x-1)^2}{a^2}-\frac{(y-4)^2}{b^2}=1.$
$\displaystyle \text{Distance between the foci}=2ae$
$\displaystyle \sqrt{(6+4)^2+(4-4)^2}=2ae$
$\displaystyle 10=2a(2)$
$\displaystyle a=\frac{5}{2}$
$\displaystyle \therefore a^2=\frac{25}{4}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =\frac{25}{4}(4-1)=\frac{75}{4}$
$\displaystyle \therefore \frac{(x-1)^2}{25/4}-\frac{(y-4)^2}{75/4}=1$
$\displaystyle \frac{4(x-1)^2}{25}-\frac{4(y-4)^2}{75}=1$
$\displaystyle 12(x-1)^2-4(y-4)^2=75$
$\displaystyle 12(x^2-2x+1)-4(y^2-8y+16)=75$
$\displaystyle 12x^2-24x+12-4y^2+32y-64=75$
$\displaystyle \therefore 12x^2-4y^2-24x+32y-127=0.$
$\displaystyle \text{Hence, the required equation is }12x^2-4y^2-24x+32y-127=0.$
$\displaystyle \\$

$\displaystyle \text{(ii) Given, the vertices are }(-8,-1)\text{ and }(16,-1)\text{ and one focus is }(17,-1).$
$\displaystyle \text{The centre is the midpoint of the line joining the vertices.}$
$\displaystyle \therefore (h,k)=\left(\frac{-8+16}{2},\frac{-1+(-1)}{2}\right)=(4,-1)$
$\displaystyle \text{Since the vertices have the same }y\text{-coordinate, the transverse axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{(x-4)^2}{a^2}-\frac{(y+1)^2}{b^2}=1.$
$\displaystyle \text{Distance between the vertices}=2a$
$\displaystyle \sqrt{(16+8)^2+(-1+1)^2}=2a$
$\displaystyle 24=2a$
$\displaystyle a=12$
$\displaystyle \therefore a^2=144$
$\displaystyle \text{The distance between the centre }(4,-1)\text{ and the focus }(17,-1)\text{ is }ae.$
$\displaystyle ae=\sqrt{(17-4)^2+(-1+1)^2}=13$
$\displaystyle 12e=13$
$\displaystyle e=\frac{13}{12}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =144\left(\frac{169}{144}-1\right)=25$
$\displaystyle \therefore \frac{(x-4)^2}{144}-\frac{(y+1)^2}{25}=1$
$\displaystyle 25(x-4)^2-144(y+1)^2=3600$
$\displaystyle 25(x^2-8x+16)-144(y^2+2y+1)=3600$
$\displaystyle 25x^2-200x+400-144y^2-288y-144=3600$
$\displaystyle \therefore 25x^2-144y^2-200x-288y-3344=0.$
$\displaystyle \text{Hence, the required equation is }25x^2-144y^2-200x-288y-3344=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the equation of the hyperbola whose:}$
$\displaystyle \text{(iii) foci are }(4,2)\text{ and }(8,2)\text{ and eccentricity is }2;$
$\displaystyle \text{(iv) vertices are }(0,\pm7)\text{ and foci are }\left(0,\pm\frac{28}{3}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii) Given, the foci are }(4,2)\text{ and }(8,2)\text{ and }e=2.$
$\displaystyle \text{The centre is the midpoint of the line joining the foci.}$
$\displaystyle \therefore (h,k)=\left(\frac{4+8}{2},\frac{2+2}{2}\right)=(6,2)$
$\displaystyle \text{Since the foci have the same }y\text{-coordinate, the transverse axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1.$
$\displaystyle \text{Distance between the foci}=2ae$
$\displaystyle \sqrt{(8-4)^2+(2-2)^2}=2ae$
$\displaystyle 4=2a(2)$
$\displaystyle a=1$
$\displaystyle \therefore a^2=1$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =1(2^2-1)=3$
$\displaystyle \therefore \frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1$
$\displaystyle 3(x-6)^2-(y-2)^2=3$
$\displaystyle 3(x^2-12x+36)-(y^2-4y+4)=3$
$\displaystyle 3x^2-36x+108-y^2+4y-4=3$
$\displaystyle \therefore 3x^2-y^2-36x+4y+101=0.$
$\displaystyle \text{Hence, the required equation is }3x^2-y^2-36x+4y+101=0.$
$\displaystyle \\$

$\displaystyle \text{(iv) Given, the vertices are }(0,\pm7)\text{ and the foci are }\left(0,\pm\frac{28}{3}\right).$
$\displaystyle \text{The centre is }(0,0)\text{ and the transverse axis is the }y\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The vertices and foci are }(0,\pm b)\text{ and }(0,\pm be)\text{ respectively.}$
$\displaystyle b=7$
$\displaystyle \therefore b^2=49$
$\displaystyle be=\frac{28}{3}$
$\displaystyle 7e=\frac{28}{3}$
$\displaystyle e=\frac{4}{3}$
$\displaystyle a^2=b^2(e^2-1)$
$\displaystyle =49\left(\frac{16}{9}-1\right)$
$\displaystyle =49\times\frac{7}{9}=\frac{343}{9}$
$\displaystyle \therefore \frac{y^2}{49}-\frac{x^2}{343/9}=1$
$\displaystyle \frac{y^2}{49}-\frac{9x^2}{343}=1$
$\displaystyle \therefore 7y^2-9x^2=343.$
$\displaystyle \text{Hence, the required equation is }7y^2-9x^2=343.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the equation of the hyperbola whose:}$
$\displaystyle \text{(v) vertices are }(\pm6,0)\text{ and one of the directrices is }x=4;$
$\displaystyle \text{(vi) foci are }(\pm2,0)\text{ and eccentricity is }\frac{3}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v) Given, the vertices are }(\pm6,0)\text{ and one of the directrices is }x=4.$
$\displaystyle \text{The centre is }(0,0)\text{ and the transverse axis is the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The vertices are }(\pm a,0).$
$\displaystyle \therefore a=6$
$\displaystyle \therefore a^2=36$
$\displaystyle \text{The directrices are }x=\pm\frac{a}{e}.$
$\displaystyle \frac{a}{e}=4$
$\displaystyle \frac{6}{e}=4$
$\displaystyle e=\frac{3}{2}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =36\left(\frac{9}{4}-1\right)$
$\displaystyle =36\times\frac{5}{4}=45$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{36}-\frac{y^2}{45}=1.$
$\displaystyle \therefore 5x^2-4y^2=180.$
$\displaystyle \\$

$\displaystyle \text{(vi) Given, the foci are }(\pm2,0)\text{ and }e=\frac{3}{2}.$
$\displaystyle \text{The centre is }(0,0)\text{ and the transverse axis is the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The foci are }(\pm ae,0).$
$\displaystyle ae=2$
$\displaystyle a\left(\frac{3}{2}\right)=2$
$\displaystyle a=\frac{4}{3}$
$\displaystyle \therefore a^2=\frac{16}{9}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =\frac{16}{9}\left(\frac{9}{4}-1\right)$
$\displaystyle =\frac{16}{9}\times\frac{5}{4}=\frac{20}{9}$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{16/9}-\frac{y^2}{20/9}=1.$
$\displaystyle \frac{9x^2}{16}-\frac{9y^2}{20}=1$
$\displaystyle \therefore 45x^2-36y^2=80.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the eccentricity of the hyperbola, the length of whose conjugate axis is }$
$\displaystyle \frac{3}{4}\text{ of the length of its transverse axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }2a\text{ and }2b\text{ be the lengths of the transverse and conjugate axes respectively,}$
$\displaystyle \text{and let }e\text{ be the eccentricity of the hyperbola.}$
$\displaystyle \text{Given, the length of the conjugate axis }=\frac{3}{4}\times\text{(length of the transverse axis).}$
$\displaystyle \therefore 2b=\frac{3}{4}(2a)$
$\displaystyle \therefore \frac{b}{a}=\frac{3}{4}$
$\displaystyle \therefore \frac{b^2}{a^2}=\frac{9}{16}$
$\displaystyle \text{Now, }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{9}{16}}$
$\displaystyle =\sqrt{\frac{25}{16}}=\frac{5}{4}$
$\displaystyle \therefore \text{The required eccentricity is }\frac{5}{4}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the equation of the hyperbola whose:}$
$\displaystyle \text{(i) focus is }(5,2),\text{ vertex is }(4,2)\text{ and centre is }(3,2);$
$\displaystyle \text{(ii) focus is }(4,2),\text{ centre is }(6,2)\text{ and }e=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, the focus is }(5,2),\text{ the vertex is }(4,2)\text{ and the centre is }(3,2).$
$\displaystyle \text{Since the centre, vertex and focus have the same }y\text{-coordinate,}$
$\displaystyle \text{the transverse axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{(x-3)^2}{a^2}-\frac{(y-2)^2}{b^2}=1.$
$\displaystyle \text{The distance between the centre }(3,2)\text{ and the vertex }(4,2)\text{ is }a.$
$\displaystyle a=\sqrt{(4-3)^2+(2-2)^2}=1$
$\displaystyle \therefore a^2=1$
$\displaystyle \text{The distance between the centre }(3,2)\text{ and the focus }(5,2)\text{ is }ae.$
$\displaystyle ae=\sqrt{(5-3)^2+(2-2)^2}=2$
$\displaystyle 1\cdot e=2$
$\displaystyle \therefore e=2$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =1(2^2-1)=3$
$\displaystyle \therefore \frac{(x-3)^2}{1}-\frac{(y-2)^2}{3}=1$
$\displaystyle 3(x-3)^2-(y-2)^2=3.$
$\displaystyle \therefore \text{The required equation is }3(x-3)^2-(y-2)^2=3.$
$\displaystyle \\$

$\displaystyle \text{(ii) Given, the focus is }(4,2),\text{ the centre is }(6,2)\text{ and }e=2.$
$\displaystyle \text{Since the centre and focus have the same }y\text{-coordinate,}$
$\displaystyle \text{the transverse axis is parallel to the }x\text{-axis.}$
$\displaystyle \text{Hence, let the equation of the hyperbola be}$
$\displaystyle \frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{b^2}=1.$
$\displaystyle \text{The distance between the centre }(6,2)\text{ and the focus }(4,2)\text{ is }ae.$
$\displaystyle ae=\sqrt{(4-6)^2+(2-2)^2}=2$
$\displaystyle a(2)=2$
$\displaystyle \therefore a=1$
$\displaystyle \therefore a^2=1$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =1(2^2-1)=3$
$\displaystyle \therefore \frac{(x-6)^2}{1}-\frac{(y-2)^2}{3}=1$
$\displaystyle 3(x-6)^2-(y-2)^2=3.$
$\displaystyle \therefore \text{The required equation is }3(x-6)^2-(y-2)^2=3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }P\text{ is any point on a hyperbola whose axes are equal, prove that}$
$\displaystyle SP\cdot S'P=CP^2,\text{ where }S\text{ and }S'\text{ are the foci and }C\text{ is the centre.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{Since the axes are equal, }a=b.$
$\displaystyle \therefore \frac{x^2}{a^2}-\frac{y^2}{a^2}=1$
$\displaystyle \therefore x^2-y^2=a^2.$
$\displaystyle \text{Also, }b^2=a^2(e^2-1).$
$\displaystyle a^2=a^2(e^2-1)$
$\displaystyle 1=e^2-1$
$\displaystyle e^2=2$
$\displaystyle \therefore e=\sqrt{2}.$
$\displaystyle \text{Hence, the centre is }C(0,0)\text{ and the foci are}$
$\displaystyle S(\sqrt{2}a,0)\text{ and }S'(-\sqrt{2}a,0).$
$\displaystyle \text{Let }P(\alpha,\beta)\text{ be any point on the hyperbola.}$
$\displaystyle \therefore \alpha^2-\beta^2=a^2.$
$\displaystyle SP^2=(\alpha-\sqrt{2}a)^2+\beta^2$
$\displaystyle =2a^2+\alpha^2+\beta^2-2\sqrt{2}a\alpha.$
$\displaystyle S'P^2=(\alpha+\sqrt{2}a)^2+\beta^2$
$\displaystyle =2a^2+\alpha^2+\beta^2+2\sqrt{2}a\alpha.$
$\displaystyle \therefore SP^2\cdot S'P^2$
$\displaystyle =\left(2a^2+\alpha^2+\beta^2-2\sqrt{2}a\alpha\right)$
$\displaystyle \qquad\times\left(2a^2+\alpha^2+\beta^2+2\sqrt{2}a\alpha\right)$
$\displaystyle =\left(2a^2+\alpha^2+\beta^2\right)^2-8a^2\alpha^2$
$\displaystyle =4a^4+4a^2(\alpha^2+\beta^2)+(\alpha^2+\beta^2)^2-8a^2\alpha^2$
$\displaystyle =4a^2(a^2-2\alpha^2)+4a^2(\alpha^2+\beta^2)+(\alpha^2+\beta^2)^2.$
$\displaystyle \text{Since }a^2=\alpha^2-\beta^2,$
$\displaystyle SP^2\cdot S'P^2$
$\displaystyle =4a^2(\alpha^2-\beta^2-2\alpha^2)+4a^2(\alpha^2+\beta^2)$
$\displaystyle \qquad+(\alpha^2+\beta^2)^2$
$\displaystyle =-4a^2(\alpha^2+\beta^2)+4a^2(\alpha^2+\beta^2)+(\alpha^2+\beta^2)^2$
$\displaystyle =(\alpha^2+\beta^2)^2.$
$\displaystyle \text{But }CP^2=\alpha^2+\beta^2.$
$\displaystyle \therefore SP^2\cdot S'P^2=CP^4.$
$\displaystyle \therefore SP\cdot S'P=CP^2.$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In each of the following, find the equation of the hyperbola satisfying the}$
$\displaystyle \text{given conditions:}$
$\displaystyle \text{(i) vertices }(\pm2,0)\text{ and foci }(\pm3,0);$
$\displaystyle \text{(ii) vertices }(0,\pm5)\text{ and foci }(0,\pm8).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given, the vertices are }(\pm2,0)\text{ and the foci are }(\pm3,0).$
$\displaystyle \text{Since the vertices lie on the }x\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The vertices and foci are }(\pm a,0)\text{ and }(\pm ae,0)\text{ respectively.}$
$\displaystyle \therefore a=2$
$\displaystyle \therefore a^2=4$
$\displaystyle ae=3$
$\displaystyle 2e=3$
$\displaystyle \therefore e=\frac{3}{2}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =4\left(\frac{9}{4}-1\right)$
$\displaystyle =4\times\frac{5}{4}=5$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{x^2}{4}-\frac{y^2}{5}=1.$
$\displaystyle \\$

$\displaystyle \text{(ii) Given, the vertices are }(0,\pm5)\text{ and the foci are }(0,\pm8).$
$\displaystyle \text{Since the vertices lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The vertices and foci are }(0,\pm b)\text{ and }(0,\pm be)\text{ respectively.}$
$\displaystyle \therefore b=5$
$\displaystyle \therefore b^2=25$
$\displaystyle be=8$
$\displaystyle 5e=8$
$\displaystyle \therefore e=\frac{8}{5}$
$\displaystyle a^2=b^2(e^2-1)$
$\displaystyle =25\left(\frac{64}{25}-1\right)$
$\displaystyle =25\times\frac{39}{25}=39$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{25}-\frac{x^2}{39}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In each of the following, find the equation of the hyperbola satisfying the}$
$\displaystyle \text{given conditions:}$
$\displaystyle \text{(iii) vertices }(0,\pm3)\text{ and foci }(0,\pm5);$
$\displaystyle \text{(iv) foci }(\pm5,0)\text{ and length of the transverse axis }=8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii) Given, the vertices are }(0,\pm3)\text{ and the foci are }(0,\pm5).$
$\displaystyle \text{Since the vertices lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The vertices and foci are }(0,\pm b)\text{ and }(0,\pm be)\text{ respectively.}$
$\displaystyle \therefore b=3$
$\displaystyle \therefore b^2=9$
$\displaystyle be=5$
$\displaystyle 3e=5$
$\displaystyle \therefore e=\frac{5}{3}$
$\displaystyle a^2=b^2(e^2-1)$
$\displaystyle =9\left(\frac{25}{9}-1\right)$
$\displaystyle =9\times\frac{16}{9}=16$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{9}-\frac{x^2}{16}=1.$
$\displaystyle \\$

$\displaystyle \text{(iv) Given, the foci are }(\pm5,0)\text{ and the length of the transverse axis is }8.$
$\displaystyle \text{Since the foci lie on the }x\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The vertices and foci are }(\pm a,0)\text{ and }(\pm ae,0)\text{ respectively.}$
$\displaystyle 2a=8$
$\displaystyle \therefore a=4$
$\displaystyle \therefore a^2=16$
$\displaystyle ae=5$
$\displaystyle 4e=5$
$\displaystyle \therefore e=\frac{5}{4}$
$\displaystyle b^2=a^2(e^2-1)$
$\displaystyle =16\left(\frac{25}{16}-1\right)$
$\displaystyle =16\times\frac{9}{16}=9$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In each of the following, find the equation of the hyperbola satisfying the}$
$\displaystyle \text{given conditions:}$
$\displaystyle \text{(v) foci }(0,\pm13)\text{ and length of the conjugate axis }=24;$
$\displaystyle \text{(vi) foci }(\pm3\sqrt{5},0)\text{ and length of the latus rectum }=8.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v) Given, the foci are }(0,\pm13)\text{ and the length of the conjugate axis is }24.$
$\displaystyle \text{Since the foci lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The length of the conjugate axis is }2a.$
$\displaystyle 2a=24$
$\displaystyle \therefore a=12$
$\displaystyle \therefore a^2=144$
$\displaystyle \text{The foci are }(0,\pm be).$
$\displaystyle \therefore be=13$
$\displaystyle \therefore b^2e^2=169$
$\displaystyle \text{Now, }a^2=b^2(e^2-1)$
$\displaystyle =b^2e^2-b^2$
$\displaystyle \therefore 144=169-b^2$
$\displaystyle \therefore b^2=25$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{25}-\frac{x^2}{144}=1.$
$\displaystyle \\$

$\displaystyle \text{(vi) Given, the foci are }(\pm3\sqrt{5},0)\text{ and the length of the latus rectum is }8.$
$\displaystyle \text{Since the foci lie on the }x\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The length of the latus rectum is }\frac{2b^2}{a}.$
$\displaystyle \therefore \frac{2b^2}{a}=8$
$\displaystyle \therefore b^2=4a\qquad\text{... ... ... ... ... (i)}$
$\displaystyle \text{The foci are }(\pm ae,0).$
$\displaystyle \therefore ae=3\sqrt{5}$
$\displaystyle \therefore a^2e^2=45.$
$\displaystyle \text{Now, }b^2=a^2(e^2-1)$
$\displaystyle \therefore b^2=a^2e^2-a^2$
$\displaystyle \therefore 4a=45-a^2$
$\displaystyle \therefore a^2+4a-45=0$
$\displaystyle \therefore (a-5)(a+9)=0$
$\displaystyle \therefore a=5\text{ or }a=-9.$
$\displaystyle \text{Since }a>0,\text{ we have }a=5.$
$\displaystyle \therefore a^2=25$
$\displaystyle \text{From (i), }b^2=4(5)=20.$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{x^2}{25}-\frac{y^2}{20}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In each of the following, find the equation of the hyperbola satisfying the}$
$\displaystyle \text{given conditions:}$
$\displaystyle \text{(vii) foci }(\pm4,0)\text{ and length of the latus rectum }=12;$
$\displaystyle \text{(viii) vertices }(0,\pm6)\text{ and }e=\frac{5}{3}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vii) Given, the foci are }(\pm4,0)\text{ and the length of the latus rectum is }12.$
$\displaystyle \text{Since the foci lie on the }x\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$
$\displaystyle \text{The length of the latus rectum is }\frac{2b^2}{a}.$
$\displaystyle \therefore \frac{2b^2}{a}=12$
$\displaystyle \therefore b^2=6a\qquad\text{... ... ... ... ... (i)}$
$\displaystyle \text{The foci are }(\pm ae,0).$
$\displaystyle \therefore ae=4$
$\displaystyle \therefore a^2e^2=16.$
$\displaystyle \text{Now, }b^2=a^2(e^2-1)$
$\displaystyle \therefore b^2=a^2e^2-a^2$
$\displaystyle \therefore 6a=16-a^2$
$\displaystyle \therefore a^2+6a-16=0$
$\displaystyle \therefore (a+8)(a-2)=0$
$\displaystyle \therefore a=-8\text{ or }a=2.$
$\displaystyle \text{Since }a>0,\text{ we have }a=2.$
$\displaystyle \therefore a^2=4$
$\displaystyle \text{From (i), }b^2=6(2)=12.$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{x^2}{4}-\frac{y^2}{12}=1.$
$\displaystyle \\$

$\displaystyle \text{(viii) Given, the vertices are }(0,\pm6)\text{ and }e=\frac{5}{3}.$
$\displaystyle \text{Since the vertices lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The vertices are }(0,\pm b).$
$\displaystyle \therefore b=6$
$\displaystyle \therefore b^2=36.$
$\displaystyle \text{Now, }a^2=b^2(e^2-1)$
$\displaystyle =36\left[\left(\frac{5}{3}\right)^2-1\right]$
$\displaystyle =36\left(\frac{25}{9}-1\right)$
$\displaystyle =36\times\frac{16}{9}$
$\displaystyle =64.$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{36}-\frac{x^2}{64}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{In each of the following, find the equation of the hyperbola satisfying the}$
$\displaystyle \text{given conditions:}$
$\displaystyle \text{(ix) foci }(0,\pm\sqrt{10})\text{ and passing through }(2,3);$
$\displaystyle \text{(x) foci }(0,\pm\sqrt{12})\text{ and length of the latus rectum }=36.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ix) Given, the foci are }(0,\pm\sqrt{10})\text{ and the hyperbola passes through }(2,3).$
$\displaystyle \text{Since the foci lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{Since }(2,3)\text{ lies on the hyperbola,}$
$\displaystyle \frac{9}{b^2}-\frac{4}{a^2}=1\qquad\text{... ... ... ... ... (i)}$
$\displaystyle \text{The foci are }(0,\pm be).$
$\displaystyle \therefore be=\sqrt{10}$
$\displaystyle \therefore b^2e^2=10.$
$\displaystyle \text{Now, }a^2=b^2(e^2-1)=b^2e^2-b^2.$
$\displaystyle \therefore a^2=10-b^2$
$\displaystyle \therefore a^2+b^2=10.\qquad\text{... ... ... ... ... (ii)}$
$\displaystyle \text{Multiplying (i) by }a^2b^2,\text{ we get}$
$\displaystyle 9a^2-4b^2=a^2b^2.$
$\displaystyle \text{Using }b^2=10-a^2,$
$\displaystyle 9a^2-4(10-a^2)=a^2(10-a^2)$
$\displaystyle 13a^2-40=10a^2-a^4$
$\displaystyle a^4+3a^2-40=0$
$\displaystyle (a^2+8)(a^2-5)=0.$
$\displaystyle \text{Since }a^2>0,\text{ we have }a^2=5.$
$\displaystyle \text{From (ii), }b^2=10-5=5.$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{5}-\frac{x^2}{5}=1.$
$\displaystyle \\$

$\displaystyle \text{(x) Given, the foci are }(0,\pm\sqrt{12})\text{ and the length of the latus rectum is }36.$
$\displaystyle \text{Since the foci lie on the }y\text{-axis, let the equation of the hyperbola be}$
$\displaystyle \frac{y^2}{b^2}-\frac{x^2}{a^2}=1.$
$\displaystyle \text{The length of the latus rectum is }\frac{2a^2}{b}.$
$\displaystyle \therefore \frac{2a^2}{b}=36$
$\displaystyle \therefore a^2=18b.\qquad\text{... ... ... ... ... (i)}$
$\displaystyle \text{The foci are }(0,\pm be).$
$\displaystyle \therefore be=\sqrt{12}$
$\displaystyle \therefore b^2e^2=12.$
$\displaystyle \text{Now, }a^2=b^2(e^2-1)=b^2e^2-b^2.$
$\displaystyle \therefore a^2=12-b^2.$
$\displaystyle \text{Using (i),}$
$\displaystyle 18b=12-b^2$
$\displaystyle b^2+18b-12=0$
$\displaystyle b=\frac{-18\pm\sqrt{18^2+48}}{2}$
$\displaystyle =\frac{-18\pm\sqrt{372}}{2}$
$\displaystyle =-9\pm\sqrt{93}.$
$\displaystyle \text{Since }b>0,\text{ we have }b=\sqrt{93}-9.$
$\displaystyle \therefore b^2=(\sqrt{93}-9)^2$
$\displaystyle =174-18\sqrt{93}.$
$\displaystyle \text{From (i),}$
$\displaystyle a^2=18(\sqrt{93}-9).$
$\displaystyle \therefore \text{The equation of the required hyperbola is}$
$\displaystyle \frac{y^2}{174-18\sqrt{93}}-\frac{x^2}{18(\sqrt{93}-9)}=1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If the distance between the foci of a hyperbola is }16\text{ and its}$
$\displaystyle \text{eccentricity is }\sqrt{2},\text{ obtain its equation.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }e=\sqrt{2}.$
$\displaystyle \text{Distance between the foci }=16.$
$\displaystyle \therefore 2ae=16$
$\displaystyle \Rightarrow 2a(\sqrt{2})=16$
$\displaystyle \Rightarrow a=4\sqrt{2}$
$\displaystyle \Rightarrow a^2=32.$
$\displaystyle \text{Now, }b^2=a^2(e^2-1)$
$\displaystyle =32(2-1)=32.$
$\displaystyle \therefore \text{The equation of the hyperbola is}$
$\displaystyle \frac{x^2}{32}-\frac{y^2}{32}=1$
$\displaystyle \Rightarrow x^2-y^2=32.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Show that the set of all points such that the difference of their distances from}$
$\displaystyle (4,0)\text{ and }(-4,0)\text{ is always equal to }2\text{ represents a hyperbola.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }F_1(4,0)\text{ and }F_2(-4,0),\text{ and let }P(x,y)\text{ be any point of the set.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \left|PF_1-PF_2\right|=2.$
$\displaystyle \text{This is the defining property of a hyperbola whose foci are }F_1\text{ and }F_2.$
$\displaystyle \text{For this hyperbola, the centre is }(0,0)\text{ and the transverse axis lies on the }x\text{-axis.}$
$\displaystyle \text{The distance between the foci is }2ae.$
$\displaystyle 2ae=8$
$\displaystyle \therefore ae=4.$
$\displaystyle \text{Also, the constant difference of the focal distances is }2a.$
$\displaystyle 2a=2$
$\displaystyle \therefore a=1$
$\displaystyle \therefore a^2=1.$
$\displaystyle ae=4$
$\displaystyle \therefore e=4.$
$\displaystyle \text{Now, }b^2=a^2(e^2-1)$
$\displaystyle =1(16-1)=15.$
$\displaystyle \therefore \text{The equation of the locus is}$
$\displaystyle \frac{x^2}{1}-\frac{y^2}{15}=1$
$\displaystyle \Rightarrow 15x^2-y^2=15.$
$\displaystyle \text{Hence, the given set of points represents a hyperbola.}$
$\displaystyle \\$