Find the equation of hyperbola having foci S(2, 1) and S'(10, 1) and a straight line `x+y-9=0` as its tangent.

Explanation

Foci are S(2, 1), S' (10, 1).
So trnasverse axis is horizontal and has the equation y = 1.
Centre is midpoint of SS', which is (6, 1).
So, equation of hyperbola is
`((x-6)^(2))/(a^(2))-((y-1)^(2))/(b^(2))=1`
Now, tangent is `x+y-9=0.`
Product of length of perpendicular from foci on the tangent is `b^(2)`.
`therefore" "b^(2)=(|2+1-9|)/(sqrt2)xx(|10+1-9|)/(sqrt2)=((6)/(sqrt2))((2)/(sqrt2))`
`therefore" "b^(2)=6`
Distance between foci, 2ac = 8
`therefore" "ae = 4`
Now, `a^(2)+b^(2)=a^(2)e^(2)`
`rArr" "a^(2)+6=16`
`rArr" "a^(2)=10`
Therefore, equation of hyperbola is
`((x-6)^(2))/(10)-((y-1)^(2))/(6)=1`
Equation of director circle is
`(x-6)^(2)+(y-1)^(2)=a^(2)-b^(2)`
`"or "(x-6)^(2)+(y-1)^(2)=4`