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Equation for the normal to a curve

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    A curve has the equation

    x^2+3xy-2y^2+17=0

    Find an expression for dy/dx in terms of x and y. I did this part correctly: dy/dx = (2x+3y)/(4y-3x)

    Find an equation for the normal to the curve at the point (3,-2).
    I got dy/dx = 0, the m of the normal = 0, then the equation of the curve being y=3, but the answer says x=3.
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    gradient of normal is not dy/dx. Thats the gradient of the tangent.
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    Yup. Derivative is the line tangent to the curve, so the 'm' for the normal is dx/dy, or undefined.
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    (Original post by aznkid66)
    Yup. Derivative is the line tangent to the curve, so the 'm' for the normal is dx/dy, or undefined.
    indicting a vertical line through (3,-2)
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    (Original post by TenOfThem)
    indicting a vertical line through (3,-2)
    So grad is infinity
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    (Original post by pwcroberts)
    So grad is infinity
    undefined rather than infinity
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    Oh, ok. Thank you!

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