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    Hey guys I am working on the following questions and I get stuck on them These are the questions I am stuck on:

    1) The curve y=x^2 - 3x -4 crosses the x-axis at P and Q. The tangents to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS.

    2) At a particular point on the curve y= 5x^2 - 12x +1 the equation of the normal is x + 18y +c =0. Find the value of the constant c.

    So for the first one, I know when y=0, the equation will factorise to (x+1)(x-4) = 0, hence x= -1,4. Then I do dy/dx to get 2x-3. When x = -1, gradient = -5 and when x = 4, gradient = 5. Now what do I do?
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    1) Work out the two equations for the tangents and find where they intersect (R), similarly for the two normals (S), and then work out the distance between the two.

    Rather a tediously long question, IMHO.

    Edit: PS, by symmetry of the quadratic about its lower point, both the normals and tangents will intersect at x= 3/2
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    (Original post by adil12)
    Hey guys I am working on the following questions and I get stuck on them These are the questions I am stuck on:

    1) The curve y=x^2 - 3x -4 crosses the x-axis at P and Q. The tangents to the curve at P and Q meet at R. The normals to the curve at P and Q meet at S. Find the distance RS.

    2) At a particular point on the curve y= 5x^2 - 12x +1 the equation of the normal is x + 18y +c =0. Find the value of the constant c.

    So for the first one, I know when y=0, the equation will factorise to (x+1)(x-4) = 0, hence x= -1,4. Then I do dy/dx to get 2x-3. When x = -1, gradient = -5 and when x = 4, gradient = 5. Now what do I do?
    1)write the equation of tangent lines
    y-y0=m(x-x0) where (x0,y0) is coordinates of P or Q and m is the
    relating gradient.
    THen solve the equations simultaneously you will get R
    For normals use same equation but with -1/m gradient
    an solve simultaneously ->S
    THen calculate the length of vector RS
    2) From the equation of normal the gradient is -1/8
    Get the derivative of y and find that x value where the derivative
    is -1/8. Calculate the value of y at x.
    Substitue the coordinates of this point in the equation of normal
    and you will get c.
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    (Original post by ghostwalker)
    1) Work out the two equations for the tangents and find where they intersect (R), similarly for the two normals (S), and then work out the distance between the two.

    Rather a tediously long question, IMHO.

    Edit: PS, by symmetry of the quadratic about its lower point, both the normals and tangents will intersect at x= 3/2
    Ok then, so Line 1 would be y = -5x + c, when x=-1, y=0, hence 5 + c =0, c + -5.
    Line 2 would be y = 5x+c, when x=4, y=0, hence 20+c = 0, c = -20.
    So -5x-5 = 5x-20, 10x = 15, x = 1.5. When x = 1.5, y = -12.5.

    How would you do it for the normals?
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    (Original post by adil12)
    Ok then, so Line 1 would be y = -5x + c, when x=-1, y=0, hence 5 + c =0, c + -5.
    Line 2 would be y = 5x+c, when x=4, y=0, hence 20+c = 0, c = -20.
    So -5x-5 = 5x-20, 10x = 15, x = 1.5. When x = 1.5, y = -12.5.

    How would you do it for the normals?
    Same as the tangents, except you get the gradient for the normal from the fact that "gradient of tangent times gradient of normal = - 1"
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    (Original post by ghostwalker)
    Same as the tangents, except you get the gradient for the normal from the fact that "gradient of tangent times gradient of normal = - 1"
    Ok then, so gradient of normal at P = 1/5 and gradient of normal at Q = -1/5.
    y = 1/5x + c and when x= -1, y=0, hence -1/5 + c =0, c = 1/5.
    And for the second line, when x = 4, y = 0, hence y = -4/5 + c = 0, c= 4/5.
    And then 1/5x + 1/5 = -1/5x + 4/5,
    2/5x = 3/5,
    x = 1.5.
    y = 1/5(1.5) + 1/5 = 1/2,
    Distance = 1/2 + 12.5 = 13, correct?
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    (Original post by adil12)
    Distance = 1/2 + 12.5 = 13, correct?
    I'm not going to check all the arithmetic, but see attached - looks right.
    Attached Images
     
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    (Original post by ghostwalker)
    I'm not going to check all the arithmetic, but see attached - looks right.
    Yup seems right thanks for the graph How would you approach question 2?
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    (Original post by adil12)
    Yup seems right thanks for the graph How would you approach question 2?
    In a nutshell:

    Find the point on the curve where the normal has the same gradient as that straight line, then sub that point into that line to find c.

    Edit: Misread. Corrected.
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    (Original post by ghostwalker)
    In a nutshell:

    Find the point on the curve where the normal has the same gradient as that straight line, then sub that point into that line to find c.

    Edit: Misread. Corrected.
    So dy/dx would be 10x - 12, but then what would you do?

    And it's fine
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    (Original post by adil12)
    So dy/dx would be 10x - 12, but then what would you do?

    And it's fine
    Does it's fine mean you've solved it now, or is that just replying to my "edit"?
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    (Original post by ghostwalker)
    Does it's fine mean you've solved it now, or is that just replying to my "edit"?
    Oh sorry I was just replying to your edit :rolleyes:
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    (Original post by adil12)
    Oh sorry I was just replying to your edit :rolleyes:

    Here's a synopsis of how to do it, as this is going to be my last post for a while.

    So what's the gradient of your straight line which is the normal?

    Hence what's the gradient of the tangent there?

    Equate to dy/dx and solve for x to find the value of x that has that gradient for the tangent, and hence that other gradient for the normal.

    Work out what the point on the curve is.

    Sub into the equation for the normal since that point lies on the normal, and solve for c.
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    Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x?
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    (Original post by adil12)
    Sorry but once I get dy/dx, how can I find the gradient when I don't know the value of x?
    :holmes: You seem to have ignored the first two steps in my synopsis.

    Given the equation of a straight line, do you know how to find its gradient?
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    (Original post by ghostwalker)
    :holmes: You seem to have ignored the first two steps in my synopsis.

    Given the equation of a straight line, do you know how to find its gradient?
    Oh yes I do, so dy/dx of the tangent = 10x -12, but without a value of x to get the value of the gradient, how am I to get he gradient of the normal?
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    (Original post by adil12)
    Oh yes I do, so dy/dx of the tangent = 10x -12, but without a value of x to get the value of the gradient, how am I to get he gradient of the normal?
    You have the equation for the normal - it's given in the question - and that's where you get the value of the gradient for the normal from.
 
 
 
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