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 equation of its directrix, the equation of its axis, the coordinates of its vertex, the equation of its latus rectum, the length of the latus rectum, and the extremities of the 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)`