[blueconstellation]

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.

[james22]

gradient of normal is not dy/dx. Thats the gradient of the tangent.

[aznkid66]

Yup. Derivative is the line tangent to the curve, so the 'm' for the normal is dx/dy, or undefined.

[TenOfThem]

(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)

[pwcroberts]

(Original post by TenOfThem)
indicting a vertical line through (3,-2)

So grad is infinity

[TenOfThem]

(Original post by pwcroberts)
So grad is infinity

undefined rather than infinity

[blueconstellation]

Oh, ok. Thank you!