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    http://files.physicsandmathstutor.co...%20Edexcel.pdf

    How can i know that the y intercept for the line perpendicular to the tangent is (0,aq)???

    I'm sure it can't be using the general equation of a tangent to the curve because although it crosses the y-axis it doesn't cross where the perpendicular line crosses the y-axis i'm sure
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    (Original post by jakaloupe)
    http://files.physicsandmathstutor.co...%20Edexcel.pdf

    How can i know that the y intercept for the line perpendicular to the tangent is (0,aq)???

    I'm sure it can't be using the general equation of a tangent to the curve because although it crosses the y-axis it doesn't cross where the perpendicular line crosses the y-axis i'm sure
    What are you talking about?

    It's asking for a line perpendicular to the tangent which goes through R. Take the tangent at Q. You are given its equation. You are told it crosses the y-axis at point R and at the y-axis the x co-ordinate is 0. Plug x=0 into the tangent equation and rearrange for y. That should give you y=aq hence R(0,aq).

    I'm fairly certain you know what the perpendicular gradient is so just plug all this information for the perp line into y-y_1=m'(x-x_1) where m' is the perp gradient and you'll have the equation of the perpendicular line going through R.
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    (Original post by RDKGames)
    What are you talking about?

    It's asking for a line perpendicular to the tangent which goes through R. Take the tangent at Q. You are given its equation. You are told it crosses the y-axis at point R and at the y-axis the x co-ordinate is 0. Plug x=0 into the tangent equation and rearrange for y. That should give you y=aq hence R(0,aq).

    I'm fairly certain you know what the perpendicular gradient is so just plug all this information for the perp line into y-y_1=m'(x-x_1) where m' is the perp gradient and you'll have the equation of the perpendicular line going through R.
    oops i read it wrong >.> why i not read properly ;_;
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    (Original post by RDKGames)
    What are you talking about?

    It's asking for a line perpendicular to the tangent which goes through R. Take the tangent at Q. You are given its equation. You are told it crosses the y-axis at point R and at the y-axis the x co-ordinate is 0. Plug x=0 into the tangent equation and rearrange for y. That should give you y=aq hence R(0,aq).

    I'm fairly certain you know what the perpendicular gradient is so just plug all this information for the perp line into y-y_1=m'(x-x_1) where m' is the perp gradient and you'll have the equation of the perpendicular line going through R.

    How can the perpendicular line pass through the point (0,aq) when the tangent goes through the line (0,aq) ?

    They're at 90° to each other so how can they cut the y-axis at the same point?
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    (Original post by jakaloupe)
    How can the perpendicular line pass through the point (0,aq) when the tangent goes through the line (0,aq) ?

    They're at 90° to each other so how can they cut the y-axis at the same point?
    You misunderstand. The question asks you to get a line which is perpendicular to the gradient. And this line must go through (0,aq) where the tangent intercepts the y-axis. It looks like this, you are finding the equation of the purple line:
    Name:  dfdsf.PNG
Views: 41
Size:  16.4 KB
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    (Original post by RDKGames)
    You misunderstand. The question asks you to get a line which is perpendicular to the gradient. And this line must go through (0,aq) where the tangent intercepts the y-axis. It looks like this, you are finding the equation of the purple line:
    Name:  dfdsf.PNG
Views: 41
Size:  16.4 KB
    Ah ok so it doesn't have to be perpendicular to the point Q, got it. Thanks
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    (Original post by jakaloupe)
    Ah ok so it doesn't have to be perpendicular to the point Q, got it. Thanks
    A point cannot be perpendicular to anything - the concept only makes sense for two lines, a line and a plane, or two planes.
 
 
 
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