Get the most out of your questions with the extensive resources available on IDNLearn.com. Get accurate answers to your questions from our community of experts who are always ready to provide timely and relevant solutions.
Sagot :
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]
We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com provides the answers you need. Thank you for visiting, and see you next time for more valuable insights.
