Question

If the pair of perpendicular lines $4x^2 + by^2 + 2cos\theta \ xy + 12x + 2sin^2\theta y + c = 0$, $\theta \in (\frac{3\pi}{2}, 2\pi)$ intersect on x-axis then the area of triangle formed by the given pair of lines and $y = 2$ is

• A. $\sqrt{65}$
• B. $\frac{\sqrt{65}}{2}$
• C. $\frac{\sqrt{65}}{4}$
• D. $2\sqrt{65}$

Answer 1

Answer: (B)

$\frac{\sqrt{65}}{2}$

Answer 2

For a pair of lines $Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0$, perpendicularity implies $A+B=0$. Here, $A=4, B=b$, so $4+b=0 \implies b=-4$. The intersection point $(x_0, y_0)$ satisfies $Ax_0 + Hy_0 + G = 0$ and $Hx_0 + By_0 + F = 0$. Since it's on the x-axis, $y_0=0$. $4x_0 + \cos\theta(0) + 6 = 0 \implies x_0 = -\frac{3}{2}$. $\cos\theta x_0 - 4(0) + \sin^2\theta = 0 \implies \cos\theta(-\frac{3}{2}) + \sin^2\theta = 0 \implies \sin^2\theta = \frac{3}{2}\cos\theta$. Using $\sin^2\theta = 1-\cos^2\theta$, we get $1-\cos^2\theta = \frac{3}{2}\cos\theta \implies 2\cos^2\theta + 3\cos\theta - 2 = 0$. Factoring gives $(2\cos\theta - 1)(\cos\theta + 2) = 0$. Since $\cos\theta \in [-1, 1]$, $\cos\theta = \frac{1}{2}$. Given $\theta \in (\frac{3\pi}{2}, 2\pi)$, $\cos\theta = \frac{1}{2}$ is valid. Then $\sin^2\theta = 1 - (\frac{1}{2})^2 = \frac{3}{4}$. The intersection point is $(-\frac{3}{2}, 0)$. Substituting this point into the equation to find $c$: $4(-\frac{3}{2})^2 - 4(0)^2 + 2(\frac{1}{2})(-\frac{3}{2})(0) + 12(-\frac{3}{2}) + 2(\frac{3}{4})(0) + c = 0 \implies 9 - 18 + c = 0 \implies c=9$. The equation of the pair of lines is $4x^2 - 4y^2 + xy + 12x + \frac{3}{2}y + 9 = 0$. To find the intersection with $y=2$: $4x^2 - 4(2)^2 + x(2) + 12x + \frac{3}{2}(2) + 9 = 0 \implies 4x^2 - 16 + 2x + 12x + 3 + 9 = 0 \implies 4x^2 + 14x - 4 = 0 \implies 2x^2 + 7x - 2 = 0$. Let the roots be $x_1, x_2$. The base of the triangle is $|x_1 - x_2| = \frac{\sqrt{7^2 - 4(2)(-2)}}{|2|} = \frac{\sqrt{49+16}}{2} = \frac{\sqrt{65}}{2}$. The height of the triangle is the distance from $(-\frac{3}{2}, 0)$ to $y=2$, which is $|2-0|=2$. Area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{\sqrt{65}}{2} \times 2 = \frac{\sqrt{65}}{2}$.