Finding equations for Tangent and Normal to the curve

[SeekerOfDragons]

Hello all,

I'm stumped on how to start the following problem:

" Find the equations of the tangent and normal to the curve: X^2 + XY - Y^2 = -1 at P(1,2) "

I know I need to find the Derivative, plug in X (in this case 1) into the derivative equation to find M and solve from there to end up with Y - 2 = M * (X - 1)

My problem is I can't figure out how to find the derivative of the equation. If I could get it into the format Y = whatever X I could do it, but the whole XY and all I can't figure out how to get Y on one side and X on the other...

Any assistance would be appreciated.

r/

SoD

 

[BigGlenntheHeavy]

Hint: A review of implicit differentation is in order.

 

[SeekerOfDragons]

Big Glenn,

Thanks for the tip. So using implicit differentiation, I came up with:

X^2 + XY - Y^2 + 1 = 0
(d/dx)(X^2 + XY - Y^2 + 1) = 0
(d/dx)(X^2) + (d/dx)(XY) - (d/dx)(Y^2) + (d/dx)(1) = 0
2X + (X(d/dx)Y + Y(d/dx)X) - 2Y(dy/dx) + 0 = 0
2X + (X(dy/dx) + Y) - 2Y(dy/dx) = 0
(dy/dx)(X - 2Y) = -2X - Y

** dy/dx = (-2X - Y) / (X - 2Y)

Would I then plug in the X and Y values from the point (1,2) to get M which will then feed into Y - 2 = M(X - 1) coming up with the equations

Y - 2 = (4/3)(X - 1)
Y = (4/3)X - (4/3) + 2

** Y = (4/3)X + (2/3) for the Tangent

and

Y - 2 = (-3/4)(X - 1)
Y = (-3/4)X + (3/4) + 2

**Y = (-3/4)X + (11/4) for the Normal

Hopefully my math and thought process is correct...

 

[BigGlenntheHeavy]

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