Question

The lines joining the origin to the points of intersection of the line $x-y=2$ with the curve $5x^2+12xy-8y^2+8x-4y+12=0$ are equally inclined to

• A. $x^2-xy=0$
• B. $xy=0$
• C. $(x-2)(y-2)=0$
• D. none of the above

Answer 1

Answer: (B), (C)

The correct options are
B $xy=0$
C $(x-2)(y-2)=0$

Lines joining the origin to the points of intersection of the line $x-y=2$ with the curve $5x^2+12xy-8y^2+8x-4y+12=0$ is obtained by homogenising the curve equation with the line equation

$5x^2+12xy-8y^2+8x(x-y)/2-4y(x-y)/2+12\left(\right(x-y)/2{)}^2=0$ is the required homogenise equation which gives the combined equation of pair of lines passing through origin.

on simplifying it becomes,

$5x^2+12xy-8y^2+4x^2-4xy-2xy+2y^2+3x^2+3y^2-6xy=0$

$12x^2-3y^2=0$

$4x^2-y^2=0$ is the required equation of pair of lines passing through origin

These pair of lines make equal inclination with pair angular bisector of these line equations

If $a{x}^2+2hxy+c{y}^2=0$ is pair of equation of line passing through origin then pair of equation of the angular bisector of these pair of lines is obtained by

$h(x^2-y^2)=(a-b)xy$

since h=0 in our required equation

The required pair of angular bisector equation is $xy=0$

since (x-2),(y-2) are parallel to our angular bisector equation lines

Therefore our pair of lines make equal angle with the $(x-2)(y-2)=0$ pair of lines.