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 = 5x4 − x2

Answer 1

y= (-9/2)*x-5/2 or 2y+9x+5=0

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