Question

If the straight lines joining the origin and the points of intersection of the curve $5x^2+12xy-6y^2+4x-2y+3=0$ and $x+ky-1=0$ are equally inclined to the co-ordinate axis, then the value of $k$

• A. is equal to $1$
• B. is equal to $-1$
• C. is equal to $2$
• D. does not exist in the set of real numbers

Answer 1

The correct option is B is equal to $-1$

Homogenizing the curve with the help of the straight line.

$5x^2+12xy-6y^2+4x(x+ky)-2y(x+ky)+3(x+ky{)}^2=0$

$12x^2+(10+4k+6k)xy+(3k^2-2k-6)y^2=0$

Lines are equally inclined to the coordinate axes

$\therefore$ coefficient of $xy=0$

$\Rightarrow 10k+10=0\Rightarrow k=-1$