Question

If \( ax^2 + 2hxy - 2ay^2 + 3x + 15y - 9 = 0 \) represents a pair of lines intersecting at (1,1), then ah =

• A. 14
• B. -15
• C. -7
• D. 9

Hint:

For a general second-degree equation \(S(x,y)=0\) representing a pair of lines, the point of intersection \( (x_0,y_0) \) satisfies the equations \( \frac{\partial S}{\partial x_0} = 0 \) and \( \frac{\partial S}{\partial y_0} = 0 \). Substitute the given intersection point into these derived linear equations to solve for unknown coefficients.

Answer 1

The Correct Option is C

Let \( S \equiv ax^2 + 2hxy - 2ay^2 + 3x + 15y - 9 = 0 \).
The point of intersection \( (x_0, y_0) \) satisfies \( \frac{\partial S}{\partial x} = 0 \) and \( \frac{\partial S}{\partial y} = 0 \).
\( \frac{\partial S}{\partial x} = 2ax + 2hy + 3 = 0 \) \( \frac{\partial S}{\partial y} = 2hx - 4ay + 15 = 0 \) Given intersection point is (1,1).
Substitute \(x=1, y=1\): \[ 2a(1) + 2h(1) + 3 = 0 \implies 2a+2h+3=0 \cdots (I) \] \[ 2h(1) - 4a(1) + 15 = 0 \implies -4a+2h+15=0 \cdots (II) \] We have a system of equations: 1) \( 2a+2h = -3 \) 2) \( -4a+2h = -15 \) Subtract (II) from (I): \( (2a - (-4a)) + (2h-2h) = -3 - (-15) \) \( 6a = 12 \implies a=2 \).
Substitute \(a=2\) into (I): \( 2(2) + 2h = -3 \implies 4+2h=-3 \implies 2h = -7 \implies h = -7/2 \).
We need to find \(ah\): \[ ah = (2) \left(-\frac{7}{2}\right) = -7 \] This matches option (3).