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Question

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

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Solution

Let the equation of the hyperbola bex2a2y2b2=1 ...(i)Then,Distance between the foci=162ae=16 [Distance between foci=2ae]ae=8a×2=8 [e=2]a=82a2=642=32Now,b2=a2(e21)=32((2)21)=32×(21)=32Putting a2=32 and b2=32 in equation(i),we getx232y232=1Hence,the equation of the required hyperbola isx2y2=32.(ii)Let the equation of the hyperbola bex2a2y2b2=1 ...(i)Then,The length of the conjugate axis=2b2b=5 [Conjugate axis =5]b=52b2=254And the distance between foci =2ae2ae=13 [The distance between foci is 13]a2e2=1694Now,b2=a2(e21)254=a2e2a21694a2a2=1694254a2=169254a2=1444=36Putting a2=36 and b2=254 in equation(i),we getx236y2254=1x2364y225=125x2144y2=900Hence,the equation of the required hyperbola is25x2144y2=900(iii),Let the equation of the hyperbola bex2a2y2b2=1 ...(i)Then,The length of the conjugate axis =2b2b=7 [Conjugate axis=5]b=72b2=494 (ii)The required hyperbola passes through the point(3,-2)(3)2a2(2)2b2=1aa2y2494=19a21649=19a2=1+16499a2=6549a2 =49×964a2=44165Putting a2=44165 and b2=494in equation (i),we getx244165y2494=165x24414y249=165x236y2441=165x236y2=441Hence,the equaion of the required hyperbola is65x236y2=441.


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