Find the value of `lambda` if the equation `(x-1)^2+(y-2)^2=lambda(x+y+3)^2` represents a parabola. Also, find its focus, the coordinates of its vertex, the equation of its latus rectum

Explanation

We have equation `(x-1)^(2)+(y-2)^(2)=lamda(x+y+3)^(2)`
`orsqrt((x-1)^(2)+(y-2)^(2))=sqrt(2lamda)(|x+y+3|)/(sqrt(1^(2)+1^(2)))`
This represents parabola if `sqrt(2lamda)=1`
`:." "lamda=(1)/(2)`
Focus of the parabola is S (1,2) and directrix is x+y+3=0.
Axis passes through focus and it is perpendicular to directrix.
So, equation of axis is x-y+1=0.
Axis and directrix meet at B(-2,-1).
Vertex A is midpoint of BS.
`:." Vertex,A"-=((1-2)/(2),(2-1)/(2))-=(-(1)/(2),(1)/(2))`
Latus rectum line is parallel to directrix and passes through focus.
So, equation of latus rectum is x+y-3=0.
Length of latus rectum `=2xx` Distance of focus from directrix
`=2xx(|1+2+3|)/(sqrt(2))=6sqrt(2)`
Extremities of latus rectum lie at distance `(6sqrt(2))/(2)=3sqrt(2)` from the focus on the latus rectum line whose equation is x+y-3=0.
Therefore, extremities of latus rectum are
`(1pm3sqrt(2)cos135^(@),2pm3sqrt(2)sin135^(@))`
`-=(1pm3,2pm3)`
`-=(4,-1)and(-2,5)`