Question

Show that all chords of the curve $3x^2-y^2-2x-4y=0$ which subtend a right angle at the origin are concurrent. Does this result hold for the curve $3x^2-3y^2-2x+4y=0$ ? If yes , What is the point of concurrency and if not, give reason.

Answer 1

Let the chord be $lx+my=1$ and it subtends a right angle at the origin. making the curve $3x^2-y^2-2x+4y=0$ homogeneous by the help of the chord, we get,
$3x^2-y^2-2(x-2y).(lx+my)=0$
$or(3-2l)x^2-2xy(m-2l)+(4m-1)y^2=0$
Since the above equation represents a pair of perpendicular lines, we have $A+B=0$
$\therefore 3-2l+4m-1=0$
$or2l-4m-2=0orl-2m=1$
Above shows that the line $lx+my=1$ passes through the point (1, -2). Now if the curve be a circle.
$3x^2-y^2-2x+4y=0$
or $x^2+y^2-\frac{2}{3}x+\frac{4}{3}y=0$
Whose centre is $(\frac{1}{3},-\frac{2}{3})$
All the chords of this circle which subtend art angle at the origin a point on the circumference must be diameters of the circle which must pass through the centre (1/3 , -2/3). Thus the result holds in the case of circles and the point of concurrency in this case is the centre of the circle.