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    2.a) Sketch the ellipse  x^2 + \frac{y^2}{4} = 1 as shown in Fig 7.6

    b) Sketch the ellipse obtained by a horizontal translation of +3 units.

    c) Write down the equation of the new ellipse.

    d) Find the co-ordinates of the points of intersection of the line y = 2x - 4 and the new ellipse.

    I'm stuck on d, its x = 2,3 (yesterday one c was incorrect)

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    Last part says

     x = \frac {1+-\sqrt {-23}}{2}
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    (Original post by ckfeister)
    2.a) Sketch the ellipse  x^2 + \frac{y^2}{4} = 1 as shown in Fig 7.6

    b) Sketch the ellipse obtained by a horizontal translation of +3 units.

    c) Write down the equation of the new ellipse.

    d) Find the co-ordinates of the points of intersection of the line y = 2x - 4 and the new ellipse.

    I'm stuck on d, its x = 2,3 (yesterday one c was incorrect)

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    You used y=2x + 4 instead of y=2x - 4
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    (Original post by solC)
    You used y=2x + 4 instead of y=2x - 4
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    3. Find the points of intersection of the hyperbola x2 - y2 = 1 and the line y = ⅓(x + 2)

    Answer is x =1.55, -1.05, yet I get below a 0 on  b^2 - 4ac
    What did I screw up this time?
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    (Original post by ckfeister)

    3. Find the points of intersection of the hyperbola x2 - y2 = 1 and the line y = ⅓(x + 2)

    Answer is x =1.55, -1.05, yet I get below a 0 on  b^2 - 4ac
    What did I screw up this time?
    Well you screwed up on calculating the discriminant. The quadratic is correct.
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    (Original post by ckfeister)
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    3. Find the points of intersection of the hyperbola x2 - y2 = 1 and the line y = ⅓(x + 2)

    Answer is x =1.55, -1.05, yet I get below a 0 on  b^2 - 4ac
    What did I screw up this time?
    The quadratic is correct
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    (Original post by RDKGames)
    Well you screwed up on calculating the determinant. The quadratic is correct.
    i think that the discriminant would be more use in this problem...
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    (Original post by the bear)
    i think that the discriminant would be more use in this problem...
    Too much linear algebra lately
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    (Original post by RDKGames)
    Too much linear algebra lately
    hehehe i did the same thing on here a whiles back

    :teehee:
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    (Original post by RDKGames)
    Well you screwed up on calculating the discriminant. The quadratic is correct.
    (Original post by solC)
    The quadratic is correct
    thx
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    2. Find the general solutions of these equations, working in radians:

    a) tan (x - π) = √3

    b) sin (½ x + π) = 1/√2

    c) cos (½ x - π/2) = √3/2



    I'm on 2c,
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    Answer is
    c) 4π/3 + 4nπ, 2π/3 + 4nπ

    I got
    4π/3 + 4nπ
    But not
    2π/3 + 4nπ

    What am I doing wrong?
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    (Original post by ckfeister)
    ...
    You forgot that  \frac{1}{2}x-\frac{\pi}{2}=2\pi n \pm \arccos(\frac{\sqrt{3}}{2}), \forall n \in \mathbb{Z}
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    (Original post by RDKGames)
    You forgot that \arccos(\frac{\sqrt{3}}{2}) =2\pi n \pm \frac{\pi}{6}, \forall n \in \mathbb{Z}
     

  \forall n \in \mathbb{Z} I'm reading new alien mathematics here, whats is this
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    (Original post by ckfeister)
     

  \forall n \in \mathbb{Z} I'm reading new alien mathematics here, whats is this
    The domain of the RHS in order to make the equation true. Meaning the statement right before it is true for all integers n. The set of integers is denoted by \mathbb{Z} and the symbol \in means "in" from the context of sets.

    And \forall just means "for every single value of..."
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    (Original post by RDKGames)
    You forgot that \arccos(\frac{\sqrt{3}}{2}) =2\pi n \pm \frac{\pi}{6}, \forall n \in \mathbb{Z}
    It's not quite true as  0\leq \arccos x \leq \pi /2 .
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    (Original post by B_9710)
    It's not quite true as  0\leq \arccos x \leq \pi /2 .
    Hah, silly me, fixed what I meant to say.
 
 
 
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