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    The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
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    (Original post by Improvement)
    The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
    What'd dy/dx at x = pi/4? What does that tell you about the gradient of the normal? How can you then write the equation of the normal?
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    (Original post by Improvement)
    The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
    have you made any start?
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    (Original post by Improvement)
    The normal to the curve y = sec^2(x) at the point P((pie/4),2) meets the line y=x at the point Q. Find the exact co-ordinates of Q. Can anyone help with this?
    Differentiate and find the value for the gradient at that point. -1/m is th gradient of the normal. Construct an equation of the normal in the form y=mx+C. Then set mx+C equal to x to find Q.
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    (Original post by Zacken)
    What'd dy/dx at x = pi/4? What does that tell you about the gradient of the normal? How can you then write the equation of the normal?
    Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
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    (Original post by Improvement)
    Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
    So you have the equation of the line as y = something *x + c

    You know that y=x

    so x =something*x +c

    x(1- something) = c

    x = ... = y
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    (Original post by Improvement)
    Yeah I have done that, 2tanx/cosx then subbed in pi/4 to get the gradient then put it -1/m to get the normals gradient. Subbed in pi/4 and 2 to get C; but I don't know how to find the co-ordinates of where x=y
    Say the normal's equation is y=2x+1, to find the intersect with y=x, you set 2x+1=x which means x and y=-1. Do the same.
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    (Original post by Kvothe the arcane)
    Say the normal's equation is y=2x+1, to find the intersect with y=x, you set 2x+1=x which means x and y=-1. Do the same.
    Where did y=2x+1 come from?
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    Where did y=2x+1 come from?
    It was an arbitrary choice of linear equation. I can't solve your question for you but thought you might find it useful to be reminded (presumably you've done this before?) of the method.
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    (Original post by Kvothe the arcane)
    It was an arbitrary choice of linear equation. I can't solve your question for you but thought you might find it useful to be reminded (presumably you've done this before?) of the method.
    Yeah I have I just was unsure about where to go once I got to the equations y=mx+c
 
 
 
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