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    Attachment 616408Can someone help me with this question? I managed to do part i where i got
    y=xp^2 -cp^3 +c/p
    For the second part I equated the y from the original equation to the y in the equation for normal and now I'm stuck in big old mess.
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    Hello? Anyone there ?
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    (Original post by TheAdviser101)
    Hello? Anyone there ?
    The attachment isn't working for me.
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    (Original post by TheAdviser101)
    Attachment 616408Can someone help me with this question? I managed to do part i where i got
    y=xp^2 -cp^3 +c/p
    For the second part I equated the y from the original equation to the y in the equation for normal and now I'm stuck in big old mess.
    im assuming your "big old mess" is correct, did you get as far as solving a quadratic using the quadratic formula?
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    Name:  IMG_1601.PNG
Views: 24
Size:  428.6 KB
    (Original post by B_9710)
    The attachment isn't working for me.
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    (Original post by DylanJ42)
    im assuming your "big old mess" is correct, did you get as far as solving a quadratic using the quadratic formula?
    Here's what I got so far:
    Name:  image.jpg
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    (Original post by TheAdviser101)
    Here's what I got so far:
    Name:  image.jpg
Views: 25
Size:  511.3 KB
    looks okay yea, write it in the form  \displaystyle ax^2 + bx + c = 0 and solve for x (youll need the quadratic formula and it might be tricky)

    you should be expecting a quadratic here since the line crosses the curve(s) at 2 places,  \displaystyle  (cp, \frac{c}{p}) and the other coordinates they want you to find
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    (Original post by DylanJ42)
    looks okay yea, write it in the form  \displaystyle ax^2 + bx + c = 0 and solve for x (youll need the quadratic formula and it might be tricky)

    you should be expecting a quadratic here since the line crosses the curve(s) at 2 places,  \displaystyle  (cp, \frac{c}{p}) and the other coordinates they want you to find
    I got (2cp , 3cp^3 + c/p) and (-c/p^3 , -cp^3)

    Attachment 616438
    Attached Images
       
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    (Original post by TheAdviser101)
    I got (2cp , 3cp^3) and (-c/p^3 , -cp^3). Is that right? I'm not able to post my solution atm

    Attachment 616438
    you made a little slip on the LHS answer  \displaystyle x = \frac{2cp^4}{2p^3} = cp \neq 2cp , otherwise thats correct

    you should always notice in these type of questions (when it says to find the other coordinate) that one of the x values you get from the quadratic should be the original x value you were given, in this case  \displaystyle x = cp is a value from the quadratic which you used originally the find the equation of the line so it can be ignored since youre only concerned with the other coordinate
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    (Original post by DylanJ42)
    you made a little slip on the LHS answer  \displaystyle x = \frac{2cp^4}{2p^3} = cp \neq 2cp , otherwise thats correct

    you should always notice in these type of questions (when it says to find the other coordinate) that one of the x values you get from the quadratic should be the original x value you were given, in this case  \displaystyle x = cp is a value from the quadratic which you used originally the find the equation of the line so it can be ignored since youre only concerned with the other coordinate


    That last bit is not making much sense to me?
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    (Original post by TheAdviser101)
    [/b]

    That last bit is not making much sense to me?
    You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
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    (Original post by B_9710)
    You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
    thank you, I couldnt think of how to explain that well
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    (Original post by B_9710)
    You solve the quadratic equation which gives the two points of intersection of the curve and the line. But one of them is going to be the point P since that's the point on the curve that the line is normal to.
    Which quadratic equation are you referring to? Is it new equation you get from equating the Cartesian to the tangent equation?
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    (Original post by TheAdviser101)
    Which quadratic equation are you referring to? Is it new equation you get from equating the Cartesian to the tangent equation?
    yea, the quadratic you solved to get the two x values, in this question one of the x values will be x=cp since you used the point P(cp, c/p) at the start to find the equation of the normal passing through P.

    Its something you should be aware of as its a nice way to know whether the x values you find from the quadratic are correct.
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    (Original post by DylanJ42)
    yea, the quadratic you solved to get the two x values, in this question one of the x values will be x=cp since you used the point P(cp, c/p) at the start to find the equation of the normal passing through P.

    Its something you should be aware of as its a nice way to know whether the x values you find from the quadratic are correct.
    Oh snap. So were the two lines intersect are the roots of the new equation. As you already know the curve already intersect/touches at x=ct you know that it should be a solution to the new equation. Thank!
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    (Original post by TheAdviser101)
    So were the two lines intersect are the roots of the new equation
    yes exactly, perfect
 
 
 
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