## 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

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## Answers ( )

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