All chords of the curve `3x^2-y^2-2x+4y=0` which subtend a right angle at the origin, pass through the fixed point

Options

A. `(1,2)`
B. `(1,-2)`
C. `(-1,2)`
D. `(-1,-2)`

Answer

B

Explanation

Given equation of the curve is `3x^2-y^2-2x+4y=0`....(i)
Let the equation of one of the chord is `y=mx+c`
`rArr (y-mx)/(c)=1` ....(ii)
On making Eq. (i) homogeneous , we get `3x^2-y^2+(-2x+4y)((y-mx)/(c))=0`
`rArr x^2(3c+2m)+y^2(-c+4)-2xy-4mxy=0`
Which represents a pair of straight lines passing through origin. Since , the angle subtended is a right angle.
`therefore 3c+2m-c+4=0`
`rArr c=-m-2`
On substituting value of c in y `=mx+c`, we get
`y=mx-m-2`
`rArr y+2=m(x-1)`
So, all such chords pass through a fixed point `(1,-2)`.