Show that all chords of the curve `3x^2-y^2-2x+4y=0,` which subtend a right angle at the origin, pass through a fixed point. Find the coordinates of the point.

Explanation

The given curve is
`3x^(2)-y^(2)-2x+4y=0` (1)
Let `y=mx+c` be the chord of combined equations of lines joining the points of intersction of curve (1) and chord `y=mx+c`to the origin can be obtained by making the equation of curve homogeneous with the help of the equation of chord as
`3x^(2)-y^(2)-2x((y-mx)/(c))+4((y-mx)/(c))=0`
or `3cx^(2)-cy^(2)-2xy+2mx^(2)+4y^(2)-4mxy=0`
or `(3c+2m)x^(2)-2(1+2m)xy+(4-c)y^(2)=0`
As the lines represented by this pair are perpendicular to each other , we must have
Coeffcient of `x^(2)+`Coefficient of `y^(2)=0`
(1) Hence , `3c+2m+4-c=0`
or `-2=m+c`
Comparing this result with `y=mx+c` , we can see that `y=mx+c`
(2) passes though (1,-2).