Key Concepts and Formulas

• Perpendicular distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.
• Trigonometric identities: $\csc \alpha = \frac{1}{\sin \alpha}$, $\sec \alpha = \frac{1}{\cos \alpha}$, $\cot 2\alpha = \frac{\cos 2\alpha}{\sin 2\alpha}$, $\sin 2\alpha = 2 \sin \alpha \cos \alpha$, $\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha$.
• $\sin^2 \alpha + \cos^2 \alpha = 1$

Step-by-Step Solution

Step 1: Analyze the First Line and Calculate $p$

The first line is given by: $x \csc \alpha - y \sec \alpha = k \cot 2\alpha$ Our goal is to convert this equation into the standard form $Ax + By + C = 0$ to easily apply the distance formula and then find the perpendicular distance $p$ from the origin.

1. Convert to sine and cosine forms: Recall that $\csc \alpha = \frac{1}{\sin \alpha}$, $\sec \alpha = \frac{1}{\cos \alpha}$, and $\cot 2\alpha = \frac{\cos 2\alpha}{\sin 2\alpha}$. Substituting these into the equation: $\frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2\alpha}{\sin 2\alpha}$

2. Simplify the Left Hand Side (LHS): To combine the terms on the LHS, we find a common denominator, which is $\sin \alpha \cos \alpha$: $\frac{x \cos \alpha - y \sin \alpha}{\sin \alpha \cos \alpha} = \frac{k \cos 2\alpha}{\sin 2\alpha}$

3. Utilize Double Angle Identity on the Right Hand Side (RHS): Recall the double angle identity $\sin 2\alpha = 2 \sin \alpha \cos \alpha$. Substitute this into the denominator of the RHS: $\frac{x \cos \alpha - y \sin \alpha}{\sin \alpha \cos \alpha} = \frac{k \cos 2\alpha}{2 \sin \alpha \cos \alpha}$

4. Clear the denominators: Multiply both sides of the equation by $2 \sin \alpha \cos \alpha$. This eliminates the denominators and simplifies the expression. (We assume $\sin \alpha \neq 0$ and $\cos \alpha \neq 0$, otherwise the original $\csc \alpha$ or $\sec \alpha$ would be undefined). $2(x \cos \alpha - y \sin \alpha) = k \cos 2\alpha$

5. Rearrange into standard form $Ax + By + C = 0$: $2x \cos \alpha - 2y \sin \alpha - k \cos 2\alpha = 0$ Here, $A = 2\cos \alpha$, $B = -2\sin \alpha$, and $C = -k \cos 2\alpha$.

6. Calculate the perpendicular distance $p$ from the origin (0, 0): Using the formula for perpendicular distance, we have: $p = \frac{|2(0) \cos \alpha - 2(0) \sin \alpha - k \cos 2\alpha|}{\sqrt{(2\cos \alpha)^2 + (-2\sin \alpha)^2}}$ $p = \frac{|-k \cos 2\alpha|}{\sqrt{4\cos^2 \alpha + 4\sin^2 \alpha}}$ $p = \frac{|k \cos 2\alpha|}{\sqrt{4(\cos^2 \alpha + \sin^2 \alpha)}}$ Since $\cos^2 \alpha + \sin^2 \alpha = 1$, we get: $p = \frac{|k \cos 2\alpha|}{\sqrt{4}} = \frac{|k \cos 2\alpha|}{2}$ Thus, $2p = |k \cos 2\alpha|$. Squaring both sides, we get: $4p^2 = k^2 \cos^2 2\alpha$

Step 2: Analyze the Second Line and Calculate $q$

The second line is given by: $x \sin \alpha + y \cos \alpha = k \sin 2\alpha$ We want to find the perpendicular distance $q$ from the origin to this line.

1. Rearrange into standard form $Ax + By + C = 0$: $x \sin \alpha + y \cos \alpha - k \sin 2\alpha = 0$ Here, $A = \sin \alpha$, $B = \cos \alpha$, and $C = -k \sin 2\alpha$.

2. Calculate the perpendicular distance $q$ from the origin (0, 0): Using the formula for perpendicular distance, we have: $q = \frac{|(0) \sin \alpha + (0) \cos \alpha - k \sin 2\alpha|}{\sqrt{(\sin \alpha)^2 + (\cos \alpha)^2}}$ $q = \frac{|-k \sin 2\alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}}$ Since $\sin^2 \alpha + \cos^2 \alpha = 1$, we get: $q = \frac{|k \sin 2\alpha|}{\sqrt{1}} = |k \sin 2\alpha|$ Squaring both sides, we get: $q^2 = k^2 \sin^2 2\alpha$

Step 3: Relate $p$ and $q$ to find $k^2$

We have: $4p^2 = k^2 \cos^2 2\alpha$ $q^2 = k^2 \sin^2 2\alpha$ Adding these two equations, we get: $4p^2 + q^2 = k^2 \cos^2 2\alpha + k^2 \sin^2 2\alpha$ $4p^2 + q^2 = k^2 (\cos^2 2\alpha + \sin^2 2\alpha)$ Since $\cos^2 2\alpha + \sin^2 2\alpha = 1$, we have: $4p^2 + q^2 = k^2$

Common Mistakes & Tips

• Be careful with trigonometric identities and make sure you are using the correct ones.
• Remember to convert the equations into the standard form $Ax + By + C = 0$ before applying the distance formula.
• Don't forget to square both sides of the equations when necessary.

Summary

We found the perpendicular distances $p$ and $q$ from the origin to the given lines. By expressing $p^2$ and $q^2$ in terms of $k^2$, $\cos^2 2\alpha$ and $\sin^2 2\alpha$ respectively, and then adding the equations, we eliminated the trigonometric terms and obtained an expression for $k^2$ in terms of $p^2$ and $q^2$. This gave us $k^2 = 4p^2 + q^2$.

Final Answer

The final answer is \boxed{4p^2 + q^2}, which corresponds to option (A).