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One Of those moments

  1. Jul 3, 2007 #1
    when there is enough block ... here is my problem.

    a hyperbola having F(-5,0) and D(5,0) and difference of focal radii 6.

    We can use |PF - PD|= 6 or PF - PF = [tex]\pm[/tex]6.

    I apply the distance formula ..eleminate radicals and simplify.

    I get

    16X^2 -9y^2=144

    Divinding both sides by 144 I get

    The hyperbolic equation

    (x^2/9)-(y^2/16) = 1

    analysing to graph ( herein lie the problems )

    First determining the intercepts ..it is easy to see the intercepts on the x axis are 3 and -3 ( verified by makeing y 0 in the above equ.) and that making x 0 we can aslo varify that there arn't any y intercepts (-y^2/16)

    ( that I understand)

    The book goes on to say to determine the extent of the graph

    making for y we get y=[tex]\pm[/tex]4/3 [square root] of x^2 -9 and
    x = [tex]\pm[/tex] 3/4 [squre root] of y^2 + 16

    y is real if and only if |x| >= 3 thus no part of the graph lies between -3 and 3 ... in the second x is real for all y variables [tex]\Leftarrow[/tex] someone explain that please.

    this part is killing me, they said notice that y = [tex]\pm[/tex]4/3 [square root] x^2 - 9 = [tex]\pm[/tex]4/3x [square root] 1-9/x^2 . [tex]\Leftarrow[/tex] I dont get this i'm thinking they factored it and i should be able to see that of course i tried ..but i cant see how . [ block]:confused:

    they the go on to say that as the absolute value of x gets larger the square root of 1- 9/x^2 get closer to 1 and therefore |x| gets larger the graphapproaches the lines y [tex]\approx[/tex][tex]\pm[/tex] 4/3x. thus as
    |x| becomes larger the graph approaches the lines y= 4/3x and -3/4x. :grumpy:



    Thanks guys
    Orson

    p.s sorry for the bad tex
     
    Last edited: Jul 3, 2007
  2. jcsd
  3. Jul 3, 2007 #2

    HallsofIvy

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    Solve for x: Adding [itex]y^2/16[/itex] to both sides, [itex]x^2/9= 1+ y^2/16= (1/16)(16+ y^2)[/itex]. Now multiply both sides by 9 to get [itex]x^2= (9/16)(16+ y^2)[/itex]. Finally, take the square root of both sides: [itex]x= \pm (3/4)\sqrt{16+ y^2}[/itex]. Since a square is never negative, [itex]16+ y^2[/itex] is never less than 16 and so [itex]\sqrt{16+ y^2}[/itex] is never less than 4. Then x cannot be between -(3/4)(4)= -3 and (3/4)(4)= 3.

    No factoring involved. Solve for y, just as we solved for x above. Adding [itex]y^2/16[/itex] to both sides and subtracting 1 from both sides, we get [itex]y^2/16= x^2/9- 1= (1/9)(x^2- 9)[/itex]. Multiply both sides of the equation by 16 to get [itex]y^2= (16/9)(x^2- 9)[/itex]. Now take the square root of both sides to get [itex]y= \pm (4/3)\sqrt{x^2- 9}[/itex]. Finally, factor that "x2" out of the square root. [itex](x^2- 9)= x^2(1- 9/x^2)[/itex] and [itex]\sqrt{x^2}= \pm x[/itex]. We have [itex]y= \pm (4/3)x\sqrt{1- 9/x^2}[/itex].

    No need to be grumpy! Yes, for very large x, [itex]9/x^2[/itex] will be very close to 9 (and getting closer as x gets larger) so [itex]\sqrt{1- 9/x^2}[/itex] will be very close to 1 (and getting closer as x gets larger and larger). That means that [itex]y= \pm (4/3)x\sqrt{1- 9/x^2}[/itex] will be very close to [itex]y= \pm (4/3)x[/itex]. The two straight lines, y= (4/3)x and y= (-4/3)x are the "asymptotes" of this hyperbola- the hyperbola gets closer and closer to those straight lines as x gots to plus or minus infinity.

    A good way of sketching this hyperbola is: Mark the vertices (-3,0) and (3,0) on the graph. Draw the rectangle with horizontal sides from (-3,4) to (3,4), (-3,-4) to (3,-4) and vertical sides from (-3,-4) to (-3, 4), (3,-4) to (3, 4). How draw the diagonals of that rectangle, extending them as far as you please outside the rectanle. Those are the asymptotes y= (4/3)x and y= (-4/3)x. Finally, starting from (3, 0), draw a smooth curve upward and to the right getting closer and closer to the asymptote y= (4/3)x. Draw a smooth curve starting at (3,0) downward and to the right getting closer and closer to the asymptote y= (-4/3)x. That give you the right half of the hyperbola. Starting from (-3,0) draw a smooth curve upward and to the left getting closer and closer to the asymptote y= (-4/3)x and a smooth curve from (-3,0) to the left and downward getting closer and closer to the asymptote y= (4/3)x. That gives you the left half of the hyperbola.



    Suggestion- use "itex" rather than "tex" for things you want in a line with other text. Keep "tex" for formulas on a separate line. Also, put whole formulas in tex, not just individual symbols.
     
  4. Jul 3, 2007 #3
    Thanks hallofIvy great explanation. Yeah the computations were getting caught in the explanation n the book and in my head, that also interseting method for sketching .. i can also plug some appropriate values for x in in the equation for y.

    Thanks again
     
    Last edited: Jul 3, 2007
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