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12th grade calculus

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find the lines that are (a) tangent and (b) normal to the curve at the given point

x^2 + xy - y^2 = 1

  • 12th grade calculus - ,

    Take the differential...

    2xdx+ y dx + x dy -2ydy=0
    solve for dy/dx
    Then use that as m in
    y=mx + b Puttingin the x,y point, solve for b.


    then, for the normal, take the negative reciprocal of m, and again, solve for the line
    y=-1/m x + b

  • 12th grade calculus - ,

    You did not state the given point.

    Using implicit derivative I found it to be

    y' = (2x+y)/(2y-x)

    sub in the given point, that gives you the slope of the tangent.
    Now that you have the slope (m) and a given point, use the grade 9 method of finding the equation of the tangent.

    Take the negative reciprocal of your slope, and the given point to find the equation of the normal.

  • 12th grade calculus - ,

    You forgot to say what point.
    However in general we find the derivative, dy/dx and call that the slope, m, of the tangent at the point. Then plug the x and y of the point in to get b in y = m x + b

    to get the derivative
    2 x dx/dx +x dy/dx + y dx/dx - 2 y dy/dx = 0
    or
    2x + y = (2y-x) dy/dx
    dy/dx = (2x+y)/(2y-x)

  • 12th grade calculus - ,

    thank alot

  • 12th grade calculus - ,

    4) x - 4y = -31
    2x - 4y = -34

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