Question

$\text{Find the centre,eccentricity,foci and directrices of the hyperbola}\left(i\right)16x^2-9y^2+32x+36y-164=0\left(ii\right)x^2-y^2+4x=0\left(iii\right)x^2-3y^2-2x=8$

Answer 1

$\left(i\right)Wehave,16x^2-9y^2+32x+36y-164=0\Rightarrow 16x^2+32x-9y^2+36y-14=0\Rightarrow 16(x^2+2x)-9(y^2+4y)-164=0\Rightarrow 16[x^2+2x+1-1]-9[y^2-4y+4-4]-164=0\Rightarrow 16\left[\right(x+1{)}^2-1]-9[(y-9{)}^2-4]-164=0\Rightarrow 16(x+1{)}^2-16-9(y-2{)}^2+36-164=0\Rightarrow 16(x+1{)}^2-9(y-2{)}^2+20-164=0\Rightarrow 16(x+1{)}^2-9(y-2{)}^2-144=0\Rightarrow 16(x+1{)}^2-9(y-2{)}^2=144=0\Rightarrow \frac{16(x+1{)}^2}{144}-\frac{9(y_2{)}^2}{144}=1\Rightarrow \frac{(x+1{)}^2}{9}-\frac{(y_2{)}^2}{16}=1...\left(i\right)\text{Shifting the origin at(-1,2)without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by x and y,}Wehavex=X-1andy=Y+2...\left(ii\right)\text{Using these relations,equation (i)reduces to}\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,wherea=0andb=16so,Wehave,Centre:Thecoordinatesofthecentrew.r.tthenewnexesare(x=0,y=0)\therefore x=-1andy=2\left[Usingequation\right(ii\left)\right]\text{So,the coordinates of the centre w.r.t the old axes are (-1,2).}\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac{5}{3}\text{Foci : The coordinates of the foci with respect to the new axes are given by}(X=\pm ae,y=0,),ie.,(X=\pm 5,Y=0)PuttingX=\pm 5andY=0inequation\left(ii\right),wegetx=\pm 5-1andy=0+2x=4,-6andy=2\text{So,the co-ordinates of foci with respect to the old axes are given by (4,2)and (-6,2)}\text{Equation of the directix : The equations of the directrix are}X=\pm \frac{a}{e}=\pm \frac{3}{\frac{5}{3}}X=\pm \frac{9}{5}PuttingX=\pm \frac{9}{5}inequation\left(ii\right),wegetx=\pm \frac{9}{5}-1PuttingX=\pm \frac{9}{5}inequation\left(ii\right),wegetx=\pm \frac{9}{5}-1\Rightarrow x=\frac{\pm 9-5}{5}\Rightarrow x=\frac{4}{5}andx=\frac{-14}{5}\text{So,the equations of the directices w.r.t the old axes are}5x-4=0and5x+14=0\left(ii\right)x^2-y^2+4x=0Wehave,x^2-y^2+4x=0\Rightarrow x^2+4x-y^2=0\Rightarrow x^2+4x+4-4-y^2=0\Rightarrow (x+2{)}^2-y^2=4\Rightarrow \frac{(x+2{)}^2}{4}-\frac{y^2}{4}=1...(i)\text{Shifting the origin at (-2,0) without rotating the axes and denoting the new co-ordinates w.r.t.these axis by x and y ,}wehavex=X-2andy=Y...(ii)\text{Using these relations,equations~~~~(i)}reducesto\frac{X^2}{4}-\frac{Y^2}{4}=1\text{This is of form}\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1,where{a}^2=4and{b}^2=4So,wehave\text{Centre : The co-ordinates of the centre w.r.t. the new axes are(X=0,Y=0)Putting X=0 and Y=0 in equation(ii),}Wegetx=-2andy=0\text{So,the coordinates of the centre w.r.t.the old axes are (-2,0)}\text{Eccentricity : The eccentricity e is given by}e=\sqrt{1+\frac{b^2}{a^2}}e=\sqrt{1}+\frac{4}{4}=\sqrt{1+1}=\sqrt{2}\text{Foci : The coordinates of the foci w.r.t.the new axes are}(x=\pm ae,y=0)i.e.,(X=\pm 2\sqrt{2},Y=0)PuttingX=0andY=0inequation(ii),wegetx=\pm 2\sqrt{2}-2andy=0x=-2\pm \sqrt{2}-2andy=0\text{So,The coordinates of the foci w.r.t.the new axes are}(-2\pm 2\sqrt{2},0)\text{Directrices : The equations of the directices w.r.t.the new axes are}X=\pm \frac{a}{e}i.e.,x=\pm \frac{2}{\sqrt{2}}PuttingX=\pm \frac{2}{\sqrt{2}}inequation(ii),wegetX=\pm \frac{2}{\sqrt{2}}=-2\Rightarrow x+2=\pm \frac{\sqrt{2}\times \sqrt{2}}{\sqrt{2}}\Rightarrow x+2=\pm \sqrt{2}\text{So,the equations of the directrices w.r.t to the old axes are}x+2=\pm \sqrt{2}.(iii)x^2-3y^2-2x=8Wehave,x^2-3y^2-2x=8\Rightarrow x^2-2x-3y=8\Rightarrow x^2-2x+1-1-3y^2=8\Rightarrow (x-1{)}^2-1-3y^2=8\Rightarrow (x-1{)}^2-3y^2=9\Rightarrow \frac{(x-1{)}^2}{0}-\frac{3y^2}{9}=1\Rightarrow \frac{(x-1{)}^2}{0}-\frac{y^2}{9}=1...(i)\text{Shifting the origin at (1,0) without rotating the axes and denoting the new co-ordinates w.r.t.these axis by x and y ,}wehavex=X+1andy=Y...(ii)\text{Using these relations,equations(i)reduces to}reducesto\frac{x^2}{9}-\frac{y^2}{9}=1...(iii)\text{This is of form}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,wherea=9and{b}^2=3so,wehave\text{Centre : The co-ordinates of the centre w.r.t. the new axes are(X=0,Y=0)Putting X=0 and Y=0 in equation(ii),}Wegetx=1andy=0\text{So,the coordinates of the centre w.r.t.the old axes are (1,0)}\text{Eccentricity : The eccentricity e is given by}e=\sqrt{\frac{b^2}{a^2}}=\sqrt{1}+\frac{3}{9}=\sqrt{\frac{1}{3}}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}=\frac{2\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}=\frac{2\sqrt{3}}{3}\text{Foci : The coordinates of the foci w.r.t.the new axes are}(x=\pm ae,y=0),i.e.,(x=\pm 2\sqrt{3},y=0)Puttingx=\pm 2\sqrt{3}andy=0inequation(ii),wegetx=\pm 2\sqrt{3}+1andy=0\Rightarrow x=1\pm 2\sqrt{3}andy=0\text{So,the coordinates of foci w.r.t.the old axes are}(1\pm 2\sqrt{3},0)\text{Directices : The equations of the directrices w.r.t the new axes are}x=\pm \frac{a}{e}i.e,x=\pm \frac{3}{\frac{2\sqrt{3}}{3}}=\pm \frac{2}{\sqrt{3}}Puttingx=\pm \frac{9}{2\sqrt{3}}+1\Rightarrow x=\pm \frac{9}{2\sqrt{3}}$