Question

$\text{Find the equation of the hyperbola whose}\left(i\right)\text{foci are (6,4)and (-4,4) and eccentricity is 2.}\left(ii\right)\text{vertices are (-8,-1) and (16,-1) and focus is(17,-1)}\left(iii\right)\text{foci are (4,2) and (8,2) and eccentricity is 2.}\left(iv\right)\text{vertices are at}(0\pm 7)andfociat(0,\pm \frac{28}{4}).$

Answer 1

$\text{(i) The centre of the hyperbola is the mi-point of the line joining the two foci.So, the coordinates of the centre are}(\frac{6-4}{2},\frac{4+4}{2})i.e.,(1,4).\text{Let conjugate axes and let e be the ecentricity.}\text{Then,the equation of the hyperbola is}\frac{(x-1{)}^2}{a^2-\frac{(y-4{)}^2}{b^2}}=1...\left(i\right)\text{Now, distance between two foci=2ae}\Rightarrow \sqrt{(6+4{)}^2+(4-4{)}^2}=2ae[\because Foci=(6,4\left)and\right(-4,4\left)\right]\Rightarrow \sqrt{(10{)}^2}=2ae\Rightarrow 10=2ae\Rightarrow 2ae=10\Rightarrow 2a\times 2=10[\because e=2]\Rightarrow a=\frac{10}{4}\Rightarrow a=\frac{5}{2}\Rightarrow a^2=\frac{25}{4}Now,b^2=a^2(e^2-1)\Rightarrow =\frac{25}{4}(2^2-1)\Rightarrow \frac{25}{4}\times 3=\frac{75}{4}Putting{a}^2=\frac{25}{4}and{b}^2=\frac{75}{4}inequation\left(i\right)weget,\Rightarrow \frac{(x-1{)}^2}{\frac{25}{4}}-\frac{(y-4{)}^2}{\frac{75}{4}}=1\Rightarrow \frac{4(x-1{)}^2}{25}-\frac{4(y-4{)}^2}{75}=1\Rightarrow \frac{4\times 3(x-1)2-4(y-4{)}^2)}{75}=1\Rightarrow 12(x-1{)}^2-4(y-4{)}^2=75\Rightarrow 12(x^2+1-2x]-4[4y^2+16-8y]=75\Rightarrow 12x^2+12-34x-4y^2-64+32y=75\Rightarrow 12x^2-4y^2-24x+32y-52-75=0\Rightarrow 12x^2-4y^2-24x+32y-127=0\text{This is the equation of the required hyperbola.}\text{(i) The cintre of the hyperbola is the mid-point of the line joining the two foci.}\text{So,the coordinates of the centre are}(\frac{16-8}{2},\frac{-1-1}{2})i.e.,(4-1).\text{Let2a and 2b the length of transverse and cinjugate axes and let e be the eccentricity.}\text{Then,the equation of the hyperbola is}\frac{(x-4{)}^2}{a^2}-\frac{(y+1{)}^2}{b^2}=1...\left(i\right)Now,\text{The distance between two vertices=2a}\therefore \sqrt{(16+8{)}^2+(-1+1{)}^2}=2ae[\because vertices=(-8,-1\left)and\right(16,-1\left)\right]\Rightarrow 24=2a\Rightarrow a=12\Rightarrow a^2=144\text{and, the distance between the focus and vertex is =ae-a}\therefore \sqrt{(17-16{)}^2+(-1+1{)}^2}=ae-1[\because Focus=(17,-1)andvertex=(16,-1\left)\right]\Rightarrow \sqrt{1^2}=ae-a\Rightarrow ae-a=1\Rightarrow 12\times e-12=1[\because a=12]\Rightarrow 12e=a+12\Rightarrow e=\frac{13}{12}\Rightarrow e^2=\frac{169}{144}Now,b^2=a^2(e^2-1)=(12{)}^2(\frac{169}{144}-1)(\because a=12ande=\frac{13}{12})=144\times \left(\frac{169-144}{144}\right)=144\times \frac{25}{144}putting{a}^2=144and{b}^2=25inequation(i),weget\frac{(x-4{)}^2}{144}-\frac{(y+1{)}^2}{25}=1\Rightarrow \frac{25(x-4{)}^2-144(y+1{)}^2}{3600}=1\Rightarrow 25[x^2+16-8x]-144[y^2+1+2y]=3600\Rightarrow 25x^2+400-200x-144y^2-144-288y=3600\Rightarrow 25x^2-144y^2-200x-288y-3344=0\text{This the equation of the required hyperbola.}(iii)\text{The centre of the hyperbola is the mid-point of the line joining the two foci.}\text{So, the coordinates of the centre are}(\frac{4+8}{2},\frac{2+2}{2})i.e.,(6,2)\text{Let 2a and 2b be the length of transverse and conjugate axes and let e be the eccentricity.}\text{Then,the equation of the hyperbola is}\frac{(x-6{)}^2}{a^2}-\frac{(y-2{)}^2}{b^2}=1...(i)\text{Now, distance between two foci=2ae}\Rightarrow \sqrt{(8-4{)}^2+(2-2{)}^2}=2ae[\because Foci=(4,2\left)and\right(8,2\left)\right]\Rightarrow \sqrt{(4{)}^2}=2ae\Rightarrow 2ae=4\Rightarrow 2\times a\times 2=4[\because e=2]\Rightarrow a=\frac{4}{4}=1\Rightarrow a^2=1Now,b^2=a^2(e^2-1)\Rightarrow b^2=1(2^2-1\left)\right[\because e=2]\Rightarrow b^2=4-1\Rightarrow b^2=3Putting{a}^2=1and{b}^2=3inequation(i),weget\frac{(x-6{)}^2}{1}-\frac{(y-2{)}^2}{3}=1\Rightarrow \frac{3(x-6{)}^2-(y-2{)}^2}{3}=1\Rightarrow 3(x-6{)}^2-(y-2{)}^2=3\Rightarrow 3(x^2+36-12x]-[y^2+4-4y]=3\Rightarrow 3x^2+108-36x-y^2-4+4y=3\Rightarrow 3x^2-y^2-36x+4y+101=0\text{This is the equation of the required hyperbola.}(iv)\text{Since,the vertices are on y-axis,so let the equation of the reuired hyperbola is}\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1...(i\left)\text{The coordinates of its vertices and foci are}\right(0,\pm be)respectively.\therefore b=7[\because vertices=(0,\pm 7)]\Rightarrow b^2=49and,be=\frac{28}{3}\left[\begin{array}{c}\because Foci=(0,\pm \frac{28}{3})\end{array}\right]\Rightarrow 7\times e=\frac{28}{3}\Rightarrow e=\frac{4}{3}\Rightarrow e^2=\frac{16}{9}Now,a^2=b^2(e^2-1)\Rightarrow a^2=49(\frac{16}{9}-1)\Rightarrow a^2=49\times \frac{7}{9}\Rightarrow a^2=\frac{343}{9}Putting{a}^2=\frac{343}{9}and{b}^2=49inequation(i),weget\frac{x^2}{\frac{343}{9}}-\frac{y^2}{49}=-1\text{This is the equation of the required hyperbola.}$