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    Equation of curve is y=20/x
    A tangent is drawn at a point on the curve
    Show that area of the triangle formed by the tangen and coordinate axes is always 40 at any point
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    (Original post by FrostedStarzzz)
    Equation of curve is y=20/x
    A tangent is drawn at a point on the curve
    Show that area of the triangle formed by the tangen and coordinate axes is always 40 at any point
    Well, start by differentiating it (you will find it easier to express it as y=20x-1 beforehand). Then find the equation of the tangent in terms of the x-coordinate of the point where it meets the curve.
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    (Original post by TheMindGarage)
    Well, start by differentiating it (you will find it easier to express it as y=20x-1 beforehand). Then find the equation of the tangent in terms of the x-coordinate of the point where it meets the curve.
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
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    (Original post by FrostedStarzzz)
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
    It's a standard expression to differentiate. \frac{20}{x}=20x^{-1}.

    Now the question says 'at any point' so the best thing to do is to pick a general point on the curve, which would have coordinates (p, \frac{20}{p}), where p \neq 0 for obvious reasons.

    Now you want to find the equation of the line in terms of p which is tangent at that general point.

    Then, you want to see where this line cuts the coordinate axis, and express those points in terms of p again.

    Finally, you want to show that the triangle defined by the y-intercept, the origin, and the x-intercept, has an area of 40 using p and showing that it cancels at some point in the calculation, hence making it negligible.
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    (Original post by FrostedStarzzz)
    I still don’t understand I differentiate it , what you mean equal tangent in terms of x coordinate at a point where it meets the curve - what I did is 20/x=20x/x^2
    20/x is equal to 20x-1. You need to be able to see how you get this. After that, it's far easier to differentiate it.

    Once you differentiated it, let a be the x-coordinate of a point on the curve. The gradient would be equal to whatever you get when you differentiate, but replacing x with a.
 
 
 
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