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Chapter 14: Problem 1211

If the lengths of perpendicular drawn from the origin to the lines \(x \cos \alpha-y \sin \alpha=\sin 2 a \alpha\) and \(x \sin \alpha+y \cos \alpha=\cos 2 \alpha\) are \(p\) and \(q\) respectively, then \(p^{2}+q^{2}=\ldots \ldots\) (a) 4 (b) 3 (c) 2 (d) 1

Short Answer

Expert verified
\(p^2 + q^2 = 1\)

Step by step solution

01

Find the equation of the lines in terms of constants A, B, and C.

We are given the equations of the lines as: \(x \cos \alpha - y \sin \alpha = \sin 2 \alpha\) \(x \sin \alpha + y \cos \alpha = \cos 2 \alpha\) Rewrite these equations in the form \(Ax + By + C = 0\): \(x \cos \alpha - y \sin \alpha - \sin 2 \alpha = 0\) \(x \sin \alpha + y \cos \alpha - \cos 2 \alpha = 0\) Now, we can identify the coefficients for A, B, and C for both equations: Line 1: \(A_1 = \cos \alpha\), \(B_1 = -\sin \alpha\), \(C_1 = -\sin 2 \alpha\) Line 2: \(A_2 = \sin \alpha\), \(B_2 = \cos \alpha\), \(C_2 = -\cos 2 \alpha\)
02

Apply the distance formula to find p and q.

The distance formula for the distance between the origin and a line is \(d = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}}\). For the given lines, we need to find \(p\) and \(q\), which are the distances of the origin to Line 1 and Line 2, respectively. For Line 1: \(p = \frac{|\cos \alpha(0) - \sin \alpha(0) - \sin 2 \alpha|}{\sqrt{(\cos \alpha)^2 + (-\sin \alpha)^2}} = \frac{|\sin 2\alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = \frac{|\sin 2\alpha|}{1}\). For Line 2: \(q = \frac{|\sin \alpha(0) + \cos \alpha(0) - \cos 2 \alpha|}{\sqrt{(\sin \alpha)^2 + (\cos \alpha)^2}} = \frac{|\cos 2\alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = \frac{|\cos 2\alpha|}{1}\).
03

Find p^2 + q^2.

Now that we have expressions for \(p\) and \(q\), we can find \(p^2 + q^2\): \(p^2 + q^2 = (\sin^2 2\alpha) + (\cos^2 2\alpha)\) Since \(\sin^2 x + \cos^2 x = 1\), we can apply this identity: \(p^2 + q^2 = 1\) Thus, the answer is: \(p^2 + q^2 = 1\) So the correct answer is (d) 1.

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