The distance between the foci of a hyperbola is 16 and its eccentricity
is `sqrt(2)`
then equation of the hyperbola is
Text Solution
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(i) Let the equation of the hyperbola be `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`. Foci are `(pmae,0)`. Distance between foci = 2ae = 16 (given) Also, `e=sqrt2` (given) `therefore" "a=4sqrt2` We know that, `b^(2)=a^(2)(e^(2)-1)` `therefore" "b^(2)=(4sqrt2)^(2)[(sqrt2)^(2)-1]=32` So, the equation of hyperbola is `(x^(2))/(32)-(y^(2))/(32)=1` or `x^(2)-y^(2)=32` If the hyperbola is of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1`, then its equation will be `x^(2)-y^(2)=-32` (ii) Vertices are `(pm4,0)` and foci are `(pm6,0).` `therefore" "a=4 and ae=6` `rArr" "e=(6)/(4)=(3)/(2)` Now, `b^(2)=a^(2)(e^(2)-1)` `therefore" "b^(2)=16((9)/(4)-1)=20` So, the equation of hyperbola is `(x^(2))/(16)-(y^(2))/(20)=1.` (iii) Since foci `(0,pmsqrt(10))` are on y-aixs, consider the equation of hyperbola as `(x^(2))/(y^(2))-(y^(2))/(b^(2))=-1.` Also, `a^(2)=b^(2)(e^(2)-1)` `therefore" "a^(2)=b^(2)e^(2)-b^(2)=10-b^(2)` `therefore" Equation of the hyperbola becomes"` `(x^(2))/(10-b^(2))-(y^(2))/(b^(2))=-1` Since, hyperbola passes through the point (2, 3), we have `therefore" "(4)/(10-b^(2))-(9)/(b^(2))=-1` `rArr" "4b^(2)-9(10-b^(2))=-b^(2)(10-b^(2))` `rArr" "b^(4)-23b^(2)+90=0` `rArr" "(b^(2)-18)(b^(2)-5)=0` `rArr" "b^(2)=5(b^(2)=18 " not possible as " a^(2)+b^(2)=10)` `rArr" "a^(2)=10-5=5` So, the equation of hyperbola is `(x^(2))/(5)-(y^(2))/(5)=-1` or `y^(2)-x^(2)=5` (iv) According to the equation `ae-a=1 and ae+a=3` `therefore" "(e+1)/(e-1)=3 or e=2` `therefore" "a=1` So, `b^(2)-a^(2)(e^(2)-1)=3` Therefore, equation is `x^(2)-(y^(2))/(3)=1` or it can be `(x^(2))/(3)-y^(2)=-1`.
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