Question

Find the equaion of the hyperbola whose

(i) focus is (0,3), directrix is x+y-1=0 and eccentricity=2

(ii) foucs is (1,1), directrix is 3x+4y+8=0 and eccentricity=2

(iii) focus is (1,1), directrix is 2x+y=1 and 3ccentricity=$\sqrt{3}$

(iv) focus is 92,-1), firectrix is 2x+3y=1 and eccentricity =2.

(v) focus is (a,0).directrix is 2x-y+a=0 and eccentricity =$\frac{4}{3}$

(vi) focus is (2,2), directiex is x+y=9 and eccentricity =2.

Answer 1

$\left(i\right)LetS(0,3)bethefocusandP(x,y)beapointonthehyperbola.$

$\text{Draw PM perpendicular from P on the directrix.Then by definition}SP=ePM\Rightarrow S{P}^2=eP{M}^2\Rightarrow (x-0{)}^2)+(y-3{)}^2=(2{)}^2{\left[\begin{array}{c}\frac{x-y+1}{\sqrt{1^2+1^2}}\end{array}\right]}^2[\because =2]\Rightarrow x^2+y^2+9-6y=\frac{4[x+y-1{]}^2}{2}\Rightarrow x+2+y^2-6y+9=2(x+y-1{)}^2\Rightarrow x+2+y^2-6y+9=2[x^2+y^2+(-1{)}^2+2xy+2\times y\times (-1)+2\times (-1)\times x]\Rightarrow x^2+y^2-6y+9=2[x^2+y^2+1+2xy-2y-2x]\Rightarrow x^2+y^2-6y+9=2x^2+2y^2+2+4xy-4y-4x\Rightarrow 2x^2-x^2+2y^2-y^2+4xy-4x-4y+6y+2-9=0\Rightarrow x^2+y^2+4xy-4x+2y-7=0\text{This is the required equation of the hyperbola.}\left(ii\right)\text{Let S(1,1) be the focus and P(x,y)be a point on the hyperbola.}$

$\text{Draw PM perpendicular from P on the directrix .Then by definition}SP=ePM\Rightarrow S{P}^2=eP{M}^2\Rightarrow (x-1{)}^2)+(y-1{)}^2=(2{)}^2{\left[\begin{array}{c}\frac{3x+4y+8}{\sqrt{3^2+4^2}}\end{array}\right]}^2[\because =2]\Rightarrow x^2+1-2x+y^2+1-2y=4\left[\begin{array}{c}\frac{3x+4y+8}{\sqrt{2}5}\end{array}\right]\Rightarrow x^2+y^2-2x-2y+2=\frac{4(2x+4y+8{)}^2}{25}\Rightarrow 23x^2+25y^2-50x-50y+50=4[9x^2+16y^2+6y+24xy+64y+48x]\Rightarrow 25x^2+25y^2-50x-50y+50=36x^2+64y^{2}56+96xy+256y+192x\Rightarrow 36x^2-25x^2+64y^2-25y^2+96xy+192x+50x+256y+50y+256-50=0\Rightarrow 11x^2+39y^2+96xy+242x+306y+206=0\text{This is the required equation of the hyperbola .}\left(iii\right)LetS(1,1)bethefocusandP(x,y)beapointonthehyperbola.$

$\text{Draw PM perpendicular from P on the directrix.Then by definition}SP=ePM\Rightarrow S{P}^2=e^{2}P{M}^2\Rightarrow (x-1{)}^2)+(y-1{)}^2=(\sqrt{3}{)}^2{\left[\begin{array}{c}\frac{2x-y-1}{\sqrt{2^2+2^2}}\end{array}\right]}^2[\because =2]\Rightarrow x^2+1-2x+y^2+1-2y=\frac{3[2x-y-1{]}^2}{5}\Rightarrow 5[x^2+y^2-2x-2y+2]=3(2x+y-1{)}^2\Rightarrow 5x^2+5y^2-10x-10y+10=3[(2x{)}^2+y^2+(-1{)}^2+2\times 2x\times y+2\times y\times (-1)+2\times (-1)\times 2x]\Rightarrow 5x^2+5y^2-10x-10y+10=3[4x^2+y^2+1+4xy-2y-4x]\Rightarrow 5x^2+5y^2-10x-10y+10=12x^2+3y^2+3+12xy-6y-12x\Rightarrow 12x^2-5x^2+3y^2-5y^2+12xy-12x+10x-6y+10y+3-10=0\Rightarrow 7x^2-2y^2+12xy-2x+4y-7=0\text{This is the required equation of the hyperbola.}(iv\left)LetS\right(2,1\left)bethefocusandP\right(x,y)beapointonthehyperbola.$

$\text{Draw PM perpendicular from P on the directrix.Then by definition}SP=ePM\Rightarrow S{P}^2=e^{2}P{M}^2\Rightarrow (x-2{)}^2)+(y+1{)}^2=2^2{\left[\begin{array}{c}\frac{2x+3y-1}{\sqrt{2^2+3^2}}\end{array}\right]}^2[\because e=2]\Rightarrow x^2+4-4x+y^2+1-2y=\frac{4[2x-3y-1{]}^2}{13}\Rightarrow 13[x^2+y^2-4x+2y+5]=4(2x+3y-1{)}^2\Rightarrow 13x^2+13y^2-52x+26y+65=4[2x+3y-1{]}^2\Rightarrow 13x^2+13y^2-52x+26y+65=4\Rightarrow [(2x^2{)}^2+(3y{)}^2+(-1{)}^2+2\times 2x\times 3y+2\times 3y\times (-1)+2\times (-1)\times 2x]\Rightarrow 13x^2+13y^2-52x+26y+65=4[4x^2+9y^2+1+12xy-6y-4x]\Rightarrow 13x^2+13y^2-52x+26y+65=16x^2+36y^2+4+48xy-24y-16x\Rightarrow 16x^2-13x^2+36y^2+-13y^2+48xy-16x+52x-24y-26y+4-65=0\Rightarrow 3x^2+23y^2+48xy+36x-50y-61=0\text{This is the required equation of the hyperbola}\left(v\right)\text{Let S(a,0)be the focus and P(x,y)be a point on the hyperbola.}$

$\text{Draw PM perpendicular from P on the directrix .Then by definition}SP=ePM\Rightarrow S{P}^2=eP{M}^2\Rightarrow (x-a{)}^2+(y-0{)}^2={\left(\frac{4}{3}\right)}^2{\left[\begin{array}{c}\frac{2x-y+a}{\sqrt{2^2+(-1{)}^2}}\end{array}\right]}^2\left[\begin{array}{c}\because e=\frac{4}{3}\end{array}\right]\Rightarrow x^2+a^2-2ax+y^2=\frac{16}{9}\times \frac{[2x-y+a{]}^2}{5}\Rightarrow 45[x^2+y^2-2ax+a^2]=16[2x-y+a{]}^2\Rightarrow 45x^2+45y^2-90ax+45a^2=16[(2x{)}^2+(-y{)}^2+a^2+2x(-y)+2\times (-y)\times a+2\times a\times 2x]\Rightarrow 45x^2+45y^2-90ax+45a^2=16[4x^2+y^2+a^2-4xy-2ay+4ax]\Rightarrow 45x^2+45y^2-90x+45a^2\Rightarrow 64x^2+16y^2+16a^2-64xy-32ay+64ax\Rightarrow 64x^2-45x^2+16y^2-45y^2-64xy+64ax+90ax-32ay+16a^2-45a^2=0\Rightarrow 19x^2-29y^2-64xy+154ax-32ay-29a^2=0\text{This is the required equation of the hyperbola.}(vi)\text{Let S (2,2)be the focus and P(x,y)be a point on the hyperbola}$

$\text{Draw PM perpendicular from P on the directrix.Then by definition}SP=ePM\Rightarrow S{P}^2=e^{2}P{M}^2\Rightarrow (x-2{)}^2+(y-2{)}^2=2^2{\left[\begin{array}{c}\frac{x+y-9}{\sqrt{1^2+1^2}}\end{array}\right]}^2\left[\begin{array}{c}\because e=\frac{4}{3}\end{array}\right]\Rightarrow x^2+4-4x+y^2+4-4y=\frac{4[x+y-9{]}^2}{2}\Rightarrow x^2+y^2-4x-4y+8=2[x+y-9{]}^2\Rightarrow x^2+y^2-4x-4y+8=2[x^2+y^2+(-9{)}^2+2\times x\times y+2\times y\times (-9)+2\times (-9)\times x]\Rightarrow x^2+y^2-4x-4y+8=2[x^2+y^2+81+2xy-18y+18x]\Rightarrow x^2+y^2-4x-4y+8=[2x^2+2y^2+162+4xy+4xy-36y-36x]\Rightarrow 2x^2-x^2+2y^2-y^2+4xy-36x+4x-36y+4y+162-8=0\Rightarrow x^2+y^2+4xy-32x-32y+154=0\text{This is the required equation of the hyperbola.}$