Given focus of hyperbola is `F_(1)(1,2)` and corresponding directrix is `2x+y-1=0.`
Let P(x,y) be the point on the hyperbola.
Foot of perpendicualr from P on the directrix is M.
So, from the definition of hyperbola, we have
`F_(1)P=e."PM"`
`therefore" "sqrt((x-1)^(2)+(y-2)^(2))=sqrt3((|2x+y-1|)/(sqrt(2^(2)+1^(2))))`
Squaring and simplifying, we get equation of hyperbola as follows:
`7x^(2)+12xy-2y^(2)-2x+14y-22=0" (1)"`
Transverse axis passes through focus `F_(1)` and perpendicular to directrix.
So, equation of transverse axis is `x-2y+3=0.`
Solving transverse axis and directrix, we get `L-=(-1//5,7//5).`
Since `A_(1)` and `A_(2)` divide `F_(1)L` in the ratio `sqrt3:1` internally an externally , respectively, we have
`A_(1)-=((5-sqrt3)/(5(sqrt3+1)),(10+7sqrt3)/(5(sqrt3+1)))`
and `A_(2)-=(-(5+sqrt3)/(5(sqrt3-1)),(7sqrt3-10)/(5(sqrt3-1)))`
Centre C of hyperbola is midpoint of `A_(1)A_(2)`, which is `(-4//5,11//10)`.
`therefore" "F_(2)-=(-(13)/(5),(1)/(5))`
