Find the equation of the hyperbola with
vertices (`pm` 5,0), foci ( `pm` 7,0).

Explanation

(i) Given that, vertices=(`pm`7,0) and `a` =(`pm`5)
`therefore` (`pm` 5,0)=(`pm` 7,0)
Now ae=7`rArr5e=7`
`rArr e=7//5`
`because b^(2)=a^(2)(e^(2)-1)`
`rArr b^(2)=25(49/25-1)`
`rArr b^(2)=25((49-25)/25)`
`rArrb^(2)=24`
So, the equation of parabola is
`(x^(2))/25-(y^(2))/24=1` [`because a^(2)=25and b^(2)=24`]
(ii) Vertices=(0,`pm`7), `e=4/3`
`therefore b=7,e=4//3`
`because e^(2)=1+(a^(2))/(b^(2))`
`rArr 16/9-1=(a^(2))/49`
`rArr 7/9=(a^(2))/49rArra^(2)=343/9`
So, the equation of hyperbola is
`-(x^(2)xx9)/343(y^(2))/49=1`
`rArr-(9x^(2))/7+y^(2)=49`
`rArr 9x^(2)-7y^(2)+343=0`
(iii) Given that, foci =90,`pmsqrt10`)
`because be=sqrt10`
`rArra^(2)+b^(2)=10`
`rArra^(2)=10-b^(2)`
`therefore` Equation of the hyperbola be
`-(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Since, this hyperbola passes through the point (2,3).
`therefore -4/(a^(2))+9/(b^(2))=1`
`rArr(-4)/(10-b^(2))+9/(b^(2))=1`
`rArr (-4b^(2)+90-9b^(2))/(b^(2)(10-b^(2))`=1
`rArr-13b^(2)+90=10b^(2)+b^(4)`
`rArrb^(4)-23b^(2)+90=0`
`rArr b^(4)-18b^(2)-5b^(2)=90=0`
`rArr b^(2)(b^(2)-18)-5(b^(2)-18)=0`
`rArr(b^(2)-18)(b^(2)-5)=0`
`rArrb^(2)=18rArrb=pm3sqrt2`
or `b^(2)=5rArrb=sqrt5`
`therefore b^(2)=18" then "a^(2)=-8` [not possible]
when `a^(2)=5" then "b^(2)=5`
So, the equation of hyperbola is
`-(x^(2))/5+(y^(2))/5=1`
`rArr y^(2)-x^(2)=5`