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How do you convert #y^2=5x# to polar form?

Answer 1

Josiah Anderson

#r =5cos theta csc^2theta #

Use #(x, y) = (r cos theta, r sin theta)#

This equation represents the parabola with vertex #V(0, +-pi/2)# and focus #S(5/4, 0)#.

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Answer 2

Aaliyah Abdullah

To convert the Cartesian equation ( y^2 = 5x ) to polar form:

Express ( x ) and ( y ) in terms of ( r ) and ( \theta ) using the relationships ( x = r \cos(\theta) ) and ( y = r \sin(\theta) ).
Substitute these expressions into the given equation.
Solve for ( r ) in terms of ( \theta ).

Starting with the given Cartesian equation ( y^2 = 5x ):

Express ( x ) in terms of ( r ) and ( \theta ): [ x = \frac{r \cos(\theta)}{\sqrt{5}} ]
Substitute ( x ) into the equation: [ (r \sin(\theta))^2 = 5 \left(\frac{r \cos(\theta)}{\sqrt{5}}\right) ]
Simplify: [ r^2 \sin^2(\theta) = r \cos(\theta) ] [ r^2 = \frac{r \cos(\theta)}{\sin^2(\theta)} ] [ r = \frac{\cos(\theta)}{\sin^2(\theta)} ]

Therefore, the polar form of the equation ( y^2 = 5x ) is ( r = \frac{\cos(\theta)}{\sin^2(\theta)} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.