Answer :
To find the polar form of the equation [tex]\( 2x^2 + 2y^2 = 3x \)[/tex], we need to convert the Cartesian coordinates [tex]\((x, y)\)[/tex] to polar coordinates [tex]\((r, \theta)\)[/tex].
1. Recall the relationships between Cartesian and polar coordinates:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]
2. Substitute these relationships into the given Cartesian equation:
[tex]\[ 2(r \cos(\theta))^2 + 2(r \sin(\theta))^2 = 3(r \cos(\theta)) \][/tex]
3. Simplify the equation step-by-step:
[tex]\[ 2r^2 \cos^2(\theta) + 2r^2 \sin^2(\theta) = 3r \cos(\theta) \][/tex]
4. Use the Pythagorean identity, [tex]\(\cos^2(\theta) + \sin^2(\theta) = 1\)[/tex]:
[tex]\[ 2r^2 (\cos^2(\theta) + \sin^2(\theta)) = 3r \cos(\theta) \][/tex]
[tex]\[ 2r^2 \cdot 1 = 3r \cos(\theta) \][/tex]
[tex]\[ 2r^2 = 3r \cos(\theta) \][/tex]
5. Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{3}{2} \cos(\theta) \][/tex]
Therefore, the polar form of the equation [tex]\( 2x^2 + 2y^2 = 3x \)[/tex] is:
[tex]\[ r = \frac{3}{2} \cos(\theta) \][/tex]
The correct answer is:
[tex]\[ r = \frac{3}{2} \cos \theta \][/tex]
1. Recall the relationships between Cartesian and polar coordinates:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]
2. Substitute these relationships into the given Cartesian equation:
[tex]\[ 2(r \cos(\theta))^2 + 2(r \sin(\theta))^2 = 3(r \cos(\theta)) \][/tex]
3. Simplify the equation step-by-step:
[tex]\[ 2r^2 \cos^2(\theta) + 2r^2 \sin^2(\theta) = 3r \cos(\theta) \][/tex]
4. Use the Pythagorean identity, [tex]\(\cos^2(\theta) + \sin^2(\theta) = 1\)[/tex]:
[tex]\[ 2r^2 (\cos^2(\theta) + \sin^2(\theta)) = 3r \cos(\theta) \][/tex]
[tex]\[ 2r^2 \cdot 1 = 3r \cos(\theta) \][/tex]
[tex]\[ 2r^2 = 3r \cos(\theta) \][/tex]
5. Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \frac{3}{2} \cos(\theta) \][/tex]
Therefore, the polar form of the equation [tex]\( 2x^2 + 2y^2 = 3x \)[/tex] is:
[tex]\[ r = \frac{3}{2} \cos(\theta) \][/tex]
The correct answer is:
[tex]\[ r = \frac{3}{2} \cos \theta \][/tex]
