The curve with the equation given below is called a kampyle of Eudoxus. Find the equation of the tangent line to this curve at the point (-1

Question

The curve with the equation given below is called a kampyle of Eudoxus. Find the equation of the tangent line to this curve at the point (-1, 2). y2 = 5×4 − x2

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Mia 5 days 2021-10-08T11:49:08+00:00 1 Answer 0

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    2021-10-08T11:51:07+00:00

    Answer:

    y= (-9/2)*x-5/2  or  2y+9x+5=0

    Step-by-step explanation:

    given the following equation

    y² = 5*x⁴ – x²

    we can take the derivative with respect to x in both sides to get the slope of the tangent line , and knowing that y=f(x) we can apply the chain rule , therefore

    d(y²)/dx = d(5*x⁴ – x²)/dx

    2*y*dy/dx = 5*4*x³ – 2*x

    dy/dx = (10*x³-x)/y

    replacing values (x=-1,y=2)

    m=dy/dx = (10*(-1)³-(-1))/2 = -9/2

    now from the equation of the line

    y= m*x+h

    we know that (x=-1,y=2) belongs also to the line since it is tangent to the curve, thus

    2= (-9/2)*(-1) + h

    h = 2 – 9/2 = (-5/2)

    thus

    y= (-9/2)*x-5/2

    or

    2y+9x+5=0

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