A hyperbola opening left and right is graphed on a polar coordinate grid with a vertical directrix 1 unit to the left of the pole.

Question
Use the image to answer the question.

Which equation represents the given graph?

(1 point)
Responses

r equals Start Fraction 3 over 2 plus 3 sine theta End Fraction
Image with alt text: r equals Start Fraction 3 over 2 plus 3 sine theta End Fraction

r equals Start Fraction 3 over 2 minus 3 sine theta End Fraction
Image with alt text: r equals Start Fraction 3 over 2 minus 3 sine theta End Fraction

r equals Start Fraction 3 over 2 plus 3 cosine theta End Fraction
Image with alt text: r equals Start Fraction 3 over 2 plus 3 cosine theta End Fraction

r equals Start Fraction 3 over 2 minus 3 cosine theta End Fraction
Image with alt text: r equals Start Fraction 3 over 2 minus 3 cosine theta End Fraction
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1 answer

A hyperbola that opens left and right has a directrix located to the left of the pole (origin). In polar coordinates, the standard form for a hyperbola is given by:

\[ r = \frac{ed}{1 \mp e \cos \theta} \]

where:

  • \( e \) is the eccentricity of the hyperbola, which is greater than 1,

  • \( d \) is the distance from the pole (origin) to the directrix,

  • The sign before \( e \cos \theta \) determines the direction of the opening:

    • For left-opening hyperbolas: use the negative sign
    • For right-opening hyperbolas: use the positive sign

Given that the vertical directrix is 1 unit to the left of the pole, we know that \( d = 1 \).

Now, also, since there is a directrix on the left side, the equation for the hyperbola should involve \(-\cos \theta\). Therefore, the form becomes:

\[ r = \frac{ed}{1 - e \cos \theta} \]

Comparing this setup with the options provided:

  1. \( r = \frac{3}{2 + 3 \sin \theta} \)
  2. \( r = \frac{3}{2 - 3 \sin \theta} \)
  3. \( r = \frac{3}{2 + 3 \cos \theta} \)
  4. \( r = \frac{3}{2 - 3 \cos \theta} \)

From the analysis, we can see that the equation for a hyperbola that opens left and right must follow the pattern where the numerator is related to \( d \) (1 unit) and the denominator has a negative sign in front of \( e \cos \theta \).

Given this context, we can choose the option that fits the equation of the hyperbola we derived.

The correct answer is:

\( r = \frac{3}{2 - 3 \cos \theta} \).