Question

If the angle between the lines joining the origin to the points of intersection of \( x + 2y + \lambda = 0 \) and \( 2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0 \) is \( \dfrac{\pi}{2} \), then a value of \( \lambda \) is:

• A. \(1\)
• B. \(\dfrac{1}{2}\)
• C. \(2\)
• D. \(\dfrac{3}{2}\)

Hint:

When angle between lines from the origin is \( \frac{\pi}{2} \), use the dot product condition \( \vec{A} \cdot \vec{B} = 0 \). Substitution reduces the system to a single variable.

Answer 1

The Correct Option is A

Step 1: Understand the condition 
We are given that the angle between the lines from origin to points of intersection of: \[ L_1: x + 2y + \lambda = 0 \] and \[ L_2: 2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0 \] is \( \frac{\pi}{2} \), i.e., 90°. So, the angle between two position vectors is \(90^\circ\), implying their dot product is 0. 
Step 2: Solve the system to find intersection points 
We substitute \( x = -2y - \lambda \) from \(L_1\) into \(L_2\): \[ 2(-2y - \lambda)^2 - 2(-2y - \lambda)y + 3y^2 + 2(-2y - \lambda) - y - 1 = 0 \] Expand step by step: \[ = 2(4y^2 + 4\lambda y + \lambda^2) + 4y^2 + 2\lambda y + 3y^2 - 4y - 2\lambda - y -1 \] Simplify: \[ = 8y^2 + 8\lambda y + 2\lambda^2 + 4y^2 + 2\lambda y + 3y^2 - 4y - 2\lambda - y -1 \] \[ = (8y^2 + 4y^2 + 3y^2) + (8\lambda y + 2\lambda y) + 2\lambda^2 - 5y - 2\lambda -1 \] \[ = 15y^2 + 10\lambda y + 2\lambda^2 - 5y - 2\lambda -1 = 0 \] This is a quadratic in \(y\). Let the two points of intersection be \( P(x_1, y_1), Q(x_2, y_2) \). Then from the origin, vectors are \( \vec{OP}, \vec{OQ} \), and the condition is \( \vec{OP} \cdot \vec{OQ} = 0 \). So we use the property: \[ \vec{OP} \cdot \vec{OQ} = x_1x_2 + y_1y_2 = 0 \] Step 3: Use known identity for perpendicular chords 
Using the relation from coordinate geometry: If vectors from origin to two points are perpendicular, then the dot product of vectors = 0. Let the intersection points be \( (x_1, y_1), (x_2, y_2) \) From symmetric property: \[ \vec{OP} \cdot \vec{OQ} = 0 \Rightarrow x_1 x_2 + y_1 y_2 = 0 \] Step 4: Use product of roots from quadratic in \(y\) 
Then: \[ y_1 y_2 = \frac{c}{a} = \frac{2\lambda^2 - 2\lambda -1}{15} \] Similarly, from \( x = -2y - \lambda \), we can express \( x_1x_2 \) in terms of \( y_1, y_2 \): \[ x = -2y - \lambda \Rightarrow x_1 x_2 = ( -2y_1 - \lambda)( -2y_2 - \lambda) \Rightarrow x_1 x_2 = 4y_1y_2 + 2\lambda(y_1 + y_2) + \lambda^2 \] Now apply: \[ x_1x_2 + y_1y_2 = 0 \] Substitute: \[ 4y_1y_2 + 2\lambda(y_1 + y_2) + \lambda^2 + y_1y_2 = 0 \Rightarrow 5y_1y_2 + 2\lambda(y_1 + y_2) + \lambda^2 = 0 \] Now use: \[ y_1y_2 = \frac{2\lambda^2 - 2\lambda -1}{15}, y_1 + y_2 = \frac{-10\lambda + 5}{15} \] Substitute back: \[ 5 \cdot \frac{2\lambda^2 - 2\lambda -1}{15} + 2\lambda \cdot \frac{-10\lambda + 5}{15} + \lambda^2 = 0 \] Simplify: \[ \frac{10\lambda^2 - 10\lambda - 5 -20\lambda^2 + 10\lambda + 15\lambda^2}{15} = 0 \Rightarrow \frac{5\lambda^2 - 5}{15} = 0 \Rightarrow \lambda^2 = 1 \Rightarrow \lambda = \pm 1 \] Step 5: Final Answer 
From options, \( \lambda = 1 \) is available. \[ \boxed{1} \]