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# Equation for the normal to a curve

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1. 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.
2. gradient of normal is not dy/dx. Thats the gradient of the tangent.
3. Yup. Derivative is the line tangent to the curve, so the 'm' for the normal is dx/dy, or undefined.
4. (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)
5. (Original post by TenOfThem)
indicting a vertical line through (3,-2)
6. (Original post by pwcroberts)
undefined rather than infinity
7. Oh, ok. Thank you!

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