Answer to Question #102904 in Geometry for ffuvj

Question

Give the equation of a conic which is symmetric to the line x+2 = 0.

Answer 1

The general equation of conic section is

$Ax^2 +Bxy +Cy^2+Dx+Ey +F =0 where, A,B,C,D,E AND F are constant$

we know that as the value of constant change then the shape of conic section will also change

we have given line equation as , x+2=0 about this line we have to find the equation of conic section

take, x=-2 now when we compare it with

$A(x+2)^2 +B(x+2)y +Cy^2+D(x+2) +Ey +F =0$

$A(x^2 +2x+4)+ Bxy +2By +Cy^2 +2Dx+2D +Ey+F=0$

$Ax^2 +Bxy+ Cy^2+ (2A+2D)x + (2B+E)y + (4A+2D+F)=0$

the above equation is symmetrical about the line x+2=0