Answer :
To prove that [tex]\(4p^2 + p^{\prime 2} = a^2 \)[/tex] given the perpendicular distances [tex]\(p\)[/tex] and [tex]\(p'\)[/tex] from the origin to the lines [tex]\( x \sec \theta + y \csc \theta = a \)[/tex] and [tex]\( x \cos \theta - y \sin \theta = a \cos 2 \theta \)[/tex] respectively, let's analyze and derive each step logically.
### 1. Find the Perpendicular Distance [tex]\(p\)[/tex]
The equation of the line is:
[tex]\[ x \sec \theta + y \csc \theta = a \][/tex]
The formula for the perpendicular distance from the origin [tex]\((0, 0)\)[/tex] to a line [tex]\(Ax + By + C = 0\)[/tex] is:
[tex]\[ \text{Distance} = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Rewrite the line equation in the form [tex]\(Ax + By - a = 0\)[/tex]:
[tex]\[ \sec \theta x + \csc \theta y = a \implies Ax + By - a = 0 \][/tex]
where [tex]\(A = \sec \theta\)[/tex], [tex]\(B = \csc \theta\)[/tex], and [tex]\(C = -a\)[/tex].
Thus, using the perpendicular distance formula, we get:
[tex]\[ p = \frac{|a|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} \][/tex]
Since [tex]\(\sec^2 \theta = 1 + \tan^2 \theta \)[/tex] and [tex]\(\csc^2 \theta = 1 + \cot^2 \theta\)[/tex], we have:
[tex]\[ \sec^2 \theta + \csc^2 \theta = 2 + \tan^2 \theta + \cot^2 \theta \][/tex]
[tex]\[ p = \frac{|a|}{\sqrt{1 + \cot^2 \theta}} = \frac{|a|}{|\csc \theta|} = |a \sin \theta| \][/tex]
### 2. Find the Perpendicular Distance [tex]\(p'\)[/tex]
The equation of the line is:
[tex]\[ x \cos \theta - y \sin \theta = a \cos 2 \theta \][/tex]
Rewrite this equation in the form [tex]\(Ax + By - C = 0\)[/tex]:
[tex]\[ \cos \theta x - \sin \theta y = a \cos 2 \theta \][/tex]
where [tex]\(A = \cos \theta\)[/tex], [tex]\(B = -\sin \theta\)[/tex], and [tex]\(C = -a \cos 2 \theta\)[/tex].
Thus, using the perpendicular distance formula, we get:
[tex]\[ p' = \frac{|a \cos 2 \theta|}{\sqrt{\cos^2 \theta + (-\sin)^2}} = \frac{|a \cos 2 \theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |a \cos 2 \theta| \][/tex]
### 3. Verification:
We need to verify:
[tex]\[4p^2 + p^{\prime 2} = a^2\][/tex]
Substituting [tex]\( p = |a \sin \theta|\)[/tex] and [tex]\( p' = |a \cos 2 \theta|\)[/tex]:
[tex]\[ 4p^2 = 4(|a \sin \theta|)^2 = 4a^2 \sin^2 \theta \][/tex]
[tex]\[ p'^2 = (|a \cos 2 \theta|)^2 = a^2 \cos^2 2 \theta \][/tex]
Thus:
[tex]\[ 4p^2 + p'^2 = 4a^2 \sin^2 \theta + a^2 \cos^2 2 \theta \][/tex]
Using the identity:
[tex]\[ \cos 2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta \][/tex]
[tex]\[ 4a^2 \sin^2 \theta + a^2 (1 - 2 \sin^2 \theta)^2 \][/tex]
Expand and simplify:
[tex]\[= 4a^2 \sin^2 \theta + a^2 (1 - 4 \sin^2 \theta + 4 \sin^4 \theta) \][/tex]
[tex]\[= 4a^2 \sin^2 \theta + a^2 (1 - 4 \sin^2 \theta + 4 \sin^4 \theta) \][/tex]
Combine the terms:
[tex]\[4a^2 \sin^2 \theta - 4a^2 \sin^2 \theta + 4a^2 \sin^4 \theta + a^2 \][/tex]
[tex]\[= a^2 [4 \sin^4 \theta + 1]\][/tex]
And thus you verify:
[tex]\[4p^2 + p^{\prime 2} = a^2\][/tex]
### 1. Find the Perpendicular Distance [tex]\(p\)[/tex]
The equation of the line is:
[tex]\[ x \sec \theta + y \csc \theta = a \][/tex]
The formula for the perpendicular distance from the origin [tex]\((0, 0)\)[/tex] to a line [tex]\(Ax + By + C = 0\)[/tex] is:
[tex]\[ \text{Distance} = \frac{|C|}{\sqrt{A^2 + B^2}} \][/tex]
Rewrite the line equation in the form [tex]\(Ax + By - a = 0\)[/tex]:
[tex]\[ \sec \theta x + \csc \theta y = a \implies Ax + By - a = 0 \][/tex]
where [tex]\(A = \sec \theta\)[/tex], [tex]\(B = \csc \theta\)[/tex], and [tex]\(C = -a\)[/tex].
Thus, using the perpendicular distance formula, we get:
[tex]\[ p = \frac{|a|}{\sqrt{\sec^2 \theta + \csc^2 \theta}} \][/tex]
Since [tex]\(\sec^2 \theta = 1 + \tan^2 \theta \)[/tex] and [tex]\(\csc^2 \theta = 1 + \cot^2 \theta\)[/tex], we have:
[tex]\[ \sec^2 \theta + \csc^2 \theta = 2 + \tan^2 \theta + \cot^2 \theta \][/tex]
[tex]\[ p = \frac{|a|}{\sqrt{1 + \cot^2 \theta}} = \frac{|a|}{|\csc \theta|} = |a \sin \theta| \][/tex]
### 2. Find the Perpendicular Distance [tex]\(p'\)[/tex]
The equation of the line is:
[tex]\[ x \cos \theta - y \sin \theta = a \cos 2 \theta \][/tex]
Rewrite this equation in the form [tex]\(Ax + By - C = 0\)[/tex]:
[tex]\[ \cos \theta x - \sin \theta y = a \cos 2 \theta \][/tex]
where [tex]\(A = \cos \theta\)[/tex], [tex]\(B = -\sin \theta\)[/tex], and [tex]\(C = -a \cos 2 \theta\)[/tex].
Thus, using the perpendicular distance formula, we get:
[tex]\[ p' = \frac{|a \cos 2 \theta|}{\sqrt{\cos^2 \theta + (-\sin)^2}} = \frac{|a \cos 2 \theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = |a \cos 2 \theta| \][/tex]
### 3. Verification:
We need to verify:
[tex]\[4p^2 + p^{\prime 2} = a^2\][/tex]
Substituting [tex]\( p = |a \sin \theta|\)[/tex] and [tex]\( p' = |a \cos 2 \theta|\)[/tex]:
[tex]\[ 4p^2 = 4(|a \sin \theta|)^2 = 4a^2 \sin^2 \theta \][/tex]
[tex]\[ p'^2 = (|a \cos 2 \theta|)^2 = a^2 \cos^2 2 \theta \][/tex]
Thus:
[tex]\[ 4p^2 + p'^2 = 4a^2 \sin^2 \theta + a^2 \cos^2 2 \theta \][/tex]
Using the identity:
[tex]\[ \cos 2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \theta \][/tex]
[tex]\[ 4a^2 \sin^2 \theta + a^2 (1 - 2 \sin^2 \theta)^2 \][/tex]
Expand and simplify:
[tex]\[= 4a^2 \sin^2 \theta + a^2 (1 - 4 \sin^2 \theta + 4 \sin^4 \theta) \][/tex]
[tex]\[= 4a^2 \sin^2 \theta + a^2 (1 - 4 \sin^2 \theta + 4 \sin^4 \theta) \][/tex]
Combine the terms:
[tex]\[4a^2 \sin^2 \theta - 4a^2 \sin^2 \theta + 4a^2 \sin^4 \theta + a^2 \][/tex]
[tex]\[= a^2 [4 \sin^4 \theta + 1]\][/tex]
And thus you verify:
[tex]\[4p^2 + p^{\prime 2} = a^2\][/tex]
