Question

If the angle between the pair of lines $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$ is $\tan^{-1}\left(\frac{3}{4}\right)$ and $h$ is a positive rational number, then the point of intersection of these two lines is

• A. $(1, -1)$
• B. $\left(-\frac{1}{9}, \frac{1}{9}\right)$
• C. $(-1, 1)$
• D. $(3, 1)$

Hint:

For a pair of lines $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, use $\tan \theta = \frac{2 \sqrt{h^2 - ab}}{|a + b|}$ to find $h$. The intersection point is found by solving $\frac{\partial F}{\partial x} = 0$ and $\frac{\partial F}{\partial y} = 0$.

Answer 1

The Correct Option is C

The equation $2x^2 + 2hxy + 2y^2 - x + y - 1 = 0$ represents a pair of lines with $a = 2$, $b = 2$, $2h = 2h$, $2g = -1$, $2f = 1$, $c = -1$. The angle $\theta$ between the lines satisfies: \[ \tan \theta = \frac{2 \sqrt{h^2 - ab}}{|a + b|} = \frac{2 \sqrt{h^2 - 2 \cdot 2}}{2 + 2} = \frac{\sqrt{h^2 - 4}}{2} = \frac{3}{4} \] \[ \sqrt{h^2 - 4} = \frac{3}{2} \implies h^2 - 4 = \frac{9}{4} \implies h^2 = \frac{25}{4} \implies h = \frac{5}{2} \quad (\text{since } h>0) \] The equation becomes: \[ 2x^2 + 5xy + 2y^2 - x + y - 1 = 0 \] The point of intersection is found by solving the partial derivatives: \[ \frac{\partial F}{\partial x} = 4x + 5y - 1 = 0 \implies 4x + 5y = 1 \] \[ \frac{\partial F}{\partial y} = 5x + 4y + 1 = 0 \implies 5x + 4y = -1 \] Multiply the first by 4 and the second by 5: \[ 16x + 20y = 4 \] \[ 25x + 20y = -5 \] Subtract: \[ -9x = 9 \implies x = -1 \] Substitute $x = -1$ into $4x + 5y = 1$: \[ 4(-1) + 5y = 1 \implies -4 + 5y = 1 \implies 5y = 5 \implies y = 1 \] The intersection is $(-1, 1)$. Option (3) is correct. Options (1), (2), and (4) do not satisfy the system.