Question

If the lines joining the origin to the points of intersection of $ 2x + 3y = k $ and $ 3x^2 - xy + 3y^2 + 2x - 3y - 4 = 0 $ are at right angles, then:

• A. \( 6k^2 + 5k + 52 = 0 \)
• B. \( 6k^2 + 5k - 52 = 0 \)
• C. \( 6k^2 - 5k + 52 = 0 \)
• D. \( 6k^2 - 5k - 52 = 0 \)

Hint:

For lines from the origin to intersection points to be perpendicular, the product of the slopes should be \( -1 \). Use Vieta’s formulas for the roots of the resulting quadratic.

Answer 1

The Correct Option is D

Step 1: Substitute to find intersection points.
From \( 2x + 3y = k \), \( y = \frac{k - 2x}{3} \). Substitute into the conic:
\[ 3x^2 - x \left( \frac{k - 2x}{3} \right) + 3 \left( \frac{k - 2x}{3} \right)^2 + 2x - 3 \left( \frac{k - 2x}{3} \right) - 4 = 0. \] Multiply by 9, simplify: \( 15x^2 + (12 - 5k)x + (k^2 - 3k - 12) = 0 \).
\[ x_1 + x_2 = \frac{5k - 12}{15}, \quad x_1 x_2 = \frac{k^2 - 3k - 12}{15}. \] Step 2: Apply the right-angle condition.
Slopes: \( \frac{k - 2x_i}{3x_i} \). Product = -1:
\[ (k - 2x_1)(k - 2x_2) = -9 x_1 x_2 \quad \Rightarrow \quad 6k^2 - 5k - 52 = 0. \]