Respuesta :

Trancescient

Step-by-step explanation:

Let [tex]x = r\cos{\theta}[/tex] and [tex]y = r\sin{\theta}[/tex]. Substituting these into our equation, it becomes

[tex]\dfrac{r^2\cos^2{\theta}}{25} + \dfrac{r^2\sin^2{\theta}}{9} = 1[/tex]

or

[tex]9r^2\cos^2{\theta} + 25r^2\sin^2{\theta} = 225[/tex]

Let use the identity [tex]\cos^2{\theta} = 1 - \sin^2{\theta}[/tex] to express the equation in terms of sine only:

[tex]9r^2 - 9r^2\sin^2{\theta} + 25\sin^2{\theta} = 225[/tex]

which simplifies to

[tex]r^2(9 + 16\sin^2{\theta}) = 225[/tex]

Taking the square root and solving for r , we get the polar form of the equation as

[tex]r = \dfrac{15}{\sqrt{9 + 16\sin^2{\theta}}}[/tex]

Thepoisnousgirl

Answer:

x = r\cos{\theta}x=rcosθ and y = r\sin{\theta}y=rsinθ

\dfrac{r^2\cos^2{\theta}}{25} + \dfrac{r^2\sin^2{\theta}}{9} = 1

25

r

2

cos

2

θ

+

9

r

2

sin

2

θ

=1

or

9r^2\cos^2{\theta} + 25r^2\sin^2{\theta} = 2259r

2

cos

2

θ+25r

2

sin

2

θ=225

Let use the identity \cos^2{\theta} = 1 - \sin^2{\theta}cos

2

θ=1−sin

2

θ

9r^2 - 9r^2\sin^2{\theta} + 25\sin^2{\theta} = 2259r

2

−9r

2

sin

2

θ+25sin

2

θ=225

which simplifies to

r^2(9 + 16\sin^2{\theta}) = 225r

2

(9+16sin

2

θ)=225

Taking the square root and solving for r ,

r = \dfrac{15}{\sqrt{9 + 16\sin^2{\theta}}}r=

9+16sin

2

θ

15

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