The square of the differences of the slopes of the lines represented by the equation $ x^2\left(\sec ^2…
The square of the differences of the slopes of the lines represented by the equation
$
x^2\left(\sec ^2 \theta-\sin ^2 \theta\right)-2 x y \tan \theta+y^2 \sin ^2 \theta=0 \text { is }
$
1
2
4
8
Solution
The given equation is $x^2(\sec^2\theta - \sin^2\theta) - 2xy\tan\theta + y^2\sin^2\theta = 0$
This is in the form of $ax^2 + 2hxy + by^2 = 0$, which represents a pair of straight lines passing through the origin.
The slope of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $m = \frac{-h \pm \sqrt{h^2 - ab}}{a}$
In the given equation, $a = \sec^2\theta - \sin^2\theta$, $h = -\tan\theta$, and $b = \sin^2\theta$.
Substituting these values in the formula for $m$, we get the slopes of the lines as
$m_1 = \frac{\tan\theta + \sqrt{\tan^2\theta - (\sec^2\theta - \sin^2\theta)\sin^2\theta}}{\sec^2\theta - \sin^2\theta}$
and
$m_2 = \frac{\tan\theta - \sqrt{\tan^2\theta - (\sec^2\theta - \sin^2\theta)\sin^2\theta}}{\sec^2\theta - \sin^2\theta}$
We need to find the square of the difference of the slopes, i.e., $(m_1 - m_2)^2$.
Subtracting $m_2$ from $m_1$, we get $m_1 - m_2 = \frac{2\sqrt{\tan^2\theta - (\sec^2\theta - \sin^2\theta)\sin^2\theta}}{\sec^2\theta - \sin^2\theta}$
Squaring both sides, we get $(m_1 - m_2)^2 = \frac{4(\tan^2\theta - (\sec^2\theta - \sin^2\theta)\sin^2\theta)}{(\sec^2\theta - \sin^2\theta)^2}$
Simplifying the expression in the numerator, we get $\tan^2\theta - \sin^2\theta = \tan^2\theta - (1 - \cos^2\theta) = \cos^2\theta$
Substituting this back into the expression for $(m_1 - m_2)^2$, we get $(m_1 - m_2)^2 = \frac{4\cos^2\theta}{(\sec^2\theta - \sin^2\theta)^2}$
But $\cos^2\theta = 1 - \sin^2\theta = \sec^2\theta - 1$, so substituting this back in, we get $(m_1 - m_2)^2 = \frac{4(\sec^2\theta - 1)}{(\sec^2\theta - \sin^2\theta)^2}$
Simplifying, we get $(m_1 - m_2)^2 = 4$
So, the square of the difference of the slopes of the lines represented by the given equation is 4. Hence, the correct option is C) 4.