Question

$\text{Find the equation of the hyperbola,referred to its principal axes as axes of coordinates,in the following cases:}\left(i\right)\text{the distance between the foci =16 and eccentricity}=\sqrt{2}\left(ii\right)\text{conjugate axis is 5 and the distance between foci=13}\left(iii\right)\text{conjugate axis is 7 and passes through the point (3,-2)}$

Answer 1

$\text{Let the equation of the hyperbola be}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)Then,\text{Distance between the foci=16}\Rightarrow 2ae=16[\because Distancebetweenfoci=2ae]\Rightarrow ae=8\Rightarrow a\times \sqrt{2}=8[\because e=\sqrt{2}]\Rightarrow a=\frac{8}{\sqrt{2}}\Rightarrow a^2=\frac{64}{2}=32Now,b^2=a^2(e^2-1)=32\left(\right(\sqrt{2}{)}^2-1)=32\times (2-1)=32Putting{a}^2=32and{b}^2=32inequation\left(i\right),weget\frac{x^2}{32}-\frac{y^2}{32}=1\text{Hence,the equation of the required hyperbola is}{x}^2-y^2=32.\left(ii\right)\text{Let the equation of the hyperbola be}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)Then,\text{The length of the conjugate axis=2b}\therefore 2b=5[\because Conjugateaxis=5]\Rightarrow b=\frac{5}{2}\Rightarrow b^2=\frac{25}{4}\text{And the distance between foci =2ae}\therefore 2ae=13[\because Thedistancebetweenfociis13]\Rightarrow a^2{e}^2=\frac{169}{4}Now,b^2=a^2(e^2-1)\Rightarrow \frac{25}{4}=a^2{e}^2-a^2\Rightarrow \frac{169}{4}-a^2\Rightarrow a^2=\frac{169}{4}-\frac{25}{4}\Rightarrow a^2=\frac{169-25}{4}\Rightarrow a^2=\frac{144}{4}=36Putting{a}^2=36and{b}^2=\frac{25}{4}inequation\left(i\right),weget\frac{x^2}{36}-\frac{y^2}{\frac{25}{4}}=1\Rightarrow \frac{x^2}{36}-\frac{4y^2}{25}=1\Rightarrow 25x^2-144y^2=900\text{Hence,the equation of the required hyperbola is}25x^2-144y^2=900\left(iii\right),\text{Let the equation of the hyperbola be}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)Then,\text{The length of the conjugate axis =2b}\therefore 2b=7[\because Conjugateaxis=5]\Rightarrow b=\frac{7}{2}{b}^2=\frac{49}{4}\left(ii\right)\text{The required hyperbola passes through the point(3,-2)}\therefore \frac{(3{)}^2}{a^2}-\frac{(-2{)}^2}{b^2}=1\Rightarrow \frac{a}{a^2}-\frac{y^2}{\frac{49}{4}}=1\Rightarrow \frac{9}{a^2}-\frac{16}{49}=1\Rightarrow \frac{9}{a^2}=1+\frac{16}{49}\Rightarrow \frac{9}{a^2}=\frac{65}{49}\Rightarrow a^2=\frac{49\times 9}{64}\Rightarrow a^2=\frac{441}{65}Putting{a}^2=\frac{441}{65}and{b}^2=\frac{49}{4}inequation\left(i\right),weget\frac{x^2}{\frac{441}{65}}-\frac{y^2}{\frac{49}{4}}=1\Rightarrow \frac{65x^2}{441}-\frac{4y^2}{49}=1\Rightarrow \frac{65x^2-36y^2}{441}=1\Rightarrow 65x^2-36y^2=441\text{Hence,the equaion of the required hyperbola is}65x^2-36y^2=441.$