Question

If the chord through $(1,-2)$ cuts the curve $3x^2 - y^2 - 2x + 4y = 0$ at P and Q, then the angle subtended by PQ at the origin is

• A. $30^\circ$
• B. $45^\circ$
• C. $60^\circ$
• D. $90^\circ$

Hint:

Angle Between Lines from Homogenized Curve.
A conic intersected by a line passing through a point gives rise to a homogeneous quadratic. If the sum of the $x^2$ and $y^2$ coefficients equals zero, the angle between resulting lines is $90^\circ$.

Answer 1

The Correct Option is D

Let the chord pass through $A = (1,-2)$. Its general form is $y + 2 = m(x - 1) \Rightarrow mx - y - (m + 2) = 0$. We homogenize the given curve $3x^2 - y^2 - 2x + 4y = 0$ with this line using: \[ 1 = \frac{mx - y}{m + 2} \Rightarrow 3x^2 - y^2 + (-2x + 4y)\left( \frac{mx - y}{m + 2} \right) \] Clearing denominators and simplifying gives a homogeneous quadratic: \[ (m+6)x^2 + (4m + 2)xy + (-m - 6)y^2 = 0 \] This represents a pair of lines through the origin. For angle between them to be $90^\circ$, we require: \[ A + B = 0 \Rightarrow m+6 + (-m -6) = 0 \] Hence, angle at origin is $90^\circ$.