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Math Help - tangents and normals to curve

  1. #1
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    tangents and normals to curve

    the curve y=x^2-3x-4 crosses the x axis a P and Q. The tangets 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?

    Really struggling thanks fo any help
    Last edited by Soloman; September 29th 2008 at 11:42 AM.
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    Quote Originally Posted by Soloman View Post
    the curve y=x^2-3x-4 crosses the y axis a P and Q. The tangets 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?

    Really struggling thanks fo any help
    the graph given crosses the y-axis once. please review your question and make sure of the wording
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    sorry meant to be x axis sorry
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    Quote Originally Posted by Soloman View Post
    the curve y=x^2-3x-4 crosses the x axis a P and Q. The tangets 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?

    Really struggling thanks fo any help

    First you find x co-ordinate of P and Q by solving the equation x^2-3x-4 = 0. For checking purposed the co-ordinate of P is (-1 , 0) and Q is (4, 0). now you must find the equation of the tangents at P and Q. First differentiate \frac{dy}{dx} = 2x - 3. So the gradient of the tangent at P is 2(-1) -3 = - 5. to find the equation of the line use the formula y - y_1 = m(x-x_1) so in this case  y - 0 = -5(x- (-1)) \ \ \rightarrow \ y= -5x - 5. I'll leave finding the equation of the tangent at Q up to you.

    Now you must find the intersection point of the tangents (that is R) and then the intersection points of the normals (that is S). then find the distance between.

    Bobak
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soloman View Post
    sorry meant to be x axis sorry
    well, then, we can find P and Q by setting y = 0 and solving for x, right.

    once we have P and Q, we can find the tangent lines at P and Q, these are just lines with slope given by the derivative at these points passing through P and Q respectively. the normal lines are the lines perpendicular to those through the same points.

    once you have all the lines, you can find their intersections to get R and S. then the distance between R and S using the distance formula
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    So I found the equations for P and Q they are:
    P y=-5x-5
    Q y=5x-20

    then I found the point of intersection then what do I do?

    x=1.5 and y=-12.5
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    Quote Originally Posted by Soloman View Post
    So I found the equations for P and Q they are:
    P y=-5x-5
    Q y=5x-20

    then I found the point of intersection then what do I do?

    x=1.5 and y=-12.5
    Now you need to find the equation of the normals at P and Q. you know know that the gradient of the normal is \frac{-1}{m} where m is the gradient of the tangent. so the gradient of the normal at P is \frac{1}{5} and the equation of the normal is y= \frac{1}{5}x + \frac{1}{5}. Can you finish the question now ?

    Bobak
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  8. #8
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    why is it 1/5 for c?
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  9. #9
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    Can someone please show me how to figure it out ive been trying for ages
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    Quote Originally Posted by Soloman View Post
    Can someone please show me how to figure it out ive been trying for ages
    if two lines are perpendicular (in particular, the normal line and the tangent line are perpendicular) then the product of their slopes is -1. that means, their slopes are the negative inverses of each other

    you had the slope of one of the tangent lines to be -5, thus, the slope of the corresponding normal line is the negative inverse of that, namely 1/5
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    thanks but what bout the rest
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    Quote Originally Posted by Soloman View Post
    why is it 1/5 for c?
    To find the equation of the normals we use the formula y - y_1 = m(x-x_1) as we did before for the tangents. however the gradients of the normal is the negative reciprocal of the gradient of the tangents.

    For example, the gradient of the tangent at P is -5 so the gradient of the normal is 1/5. then using the equation with the point (-1,0) you have y - 0 = \frac{1}{5}(x-(-1)) \ \rightarrow \ y= \frac{1}{5}x + \frac{1}{5}..

    For Q the gradient of the normal is -\frac{1}{5}. can you now find the equation the normal at Q and finish off the question ?

    Bobak
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  13. #13
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    okay so I have
    p = y=1/5x+1/5
    q= -1/5x+4/5

    so now I try and find the point of intersection right by making them equal each other then that will give me the x then I put it into one of the first equations to give me y. Then I have found coordinate of R. So how do you make an equation of a line that is perpendicular to R when u only have its coordinates?
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  14. #14
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    Quote Originally Posted by Soloman View Post
    So how do you make an equation of a line that is perpendicular to R when u only have its coordinates?
    I don't understand why you are asking this, your question is to find the distance between S and R. do you know how to find the distance between two points ?

    Bobak
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  15. #15
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    okay so this is what I have for R its coordinates are x=1.5 and y=0.5
    S is the normal of R is that right?
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