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# Find equations of the parabola?

Hi, please help me with this.

Find the equations of the hyperbola, assuming form (x^2/a^2) - (y^2/b^2) = 1:

1. Distances between directrices unity, passing

through (2,3).

2. Latus rectum 6, distance between foci twice the

distance between directrices.

3. Foci (4,0)(-4,0), slope of asymptotes +3 and -3.

I only need solutions to these. The answers given

are:

1. 3x^2 - y^2 =3; 12x^2 - y^2 =39

2. x^2 -y^2 = 9

3. 45x^2 - 5y^2 = 72

Thanks in adv to those who'll answer.

### 1 Answer

- PopeLv 76 years agoFavorite Answer
Your opening condition has the center at the origin and the transverse axis is horizontal.

[1]

Distance between the directrices is unity.

Let the directices be x = -1/2 and x = 1/2.

The corresponding foci are F(-u, 0) and G(u, 0), where u > 0.

Let the given point be P(2, 3).

Let P be projected horizontally, meeting the directrices at Q(-1/2, 3) and R(1/2, 3).

PF/PQ = PG/PR

√[(2 + u)² + 3²] / (5/2) = √[(2 - u)² + 3²] / (3/2)

3√[(2 + u)² + 3²] = 5√[(2 - u)² + 3²]

9[(2 + u)² + 3²] = 25[(2 - u)² + 3²]

9(u² + 4u + 13) = 25(u² - 4u + 13)

16u² - 136u + 208 = 0

2u² - 17u + 26 = 0

(u - 2)(2u - 13) = 0

u = 2 or u = 13/2

That gives you the foci and the directrices. Can you finish that?

[2]

Let the directrices be x = -v and x = v, for v > 0.

Foci: (-2v, 0), (2v, 0)

Latus rectum is 6, so we have four points on the section.

Here is one of them: (2v, 3)

Use a proportion similar to the previous one.

√[(4v)² + 3²] / (3v) = 3 / v

√[(4v)² + 3²] = 9

16v² + 9 = 81

v² = 9/2

v = 3/√(2)

Directrices: x = -3/√(2), x = 3/√(2)

Foci: (-3√(2), 0), (3√(2), 0)

[3]

Draw vertical lines through the vertices. They intersect the asymptotes at four points, each of them a distance of 4 from the origin.

Let w be the transverse radius, That makes 3w the conjugate radius.

w² + (3w)² = 4²

10w² = 16

w = √(8/5)

x²/(8/5) - y²/(72/5) = 1

45x² - 5y² - 72 = 0

I felt like I had to finish one of them.

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