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Find the equation of the hyperbola with foci (±4,0) and length of the latus rectum is 12 |
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Answer» Answer:3x^2-y^2=12Step-by-step explanation:LET equation of HYPERBOLA bex^2/a^2-y^2-b^2=1F(+0,0) = F(+4,0)THEREFORE, c=4 and c^2= 1612=lr(length)=2b^2/a= b^2= 6ac^2= a^2+ b^216-a^2+ 6aa^2+ 6a-16=0Therefore, a=- 8, 2 but a cannot be negative.a=2 Therefore, a^2= 4AND b^2=6a= 6x2=12put a^2 and b^2 value in equationx^2/ 4- y^2/ 12= 13x^2- y^2= 12 |
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