The curve with the equation y^2=5x^4-x^2 is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point (-1,2).

Answer 1

y^2=5x^4-x^2

2y dy/dx = 20x^3 - 2x
dy/dx = (20x^3 - 2x)/(2y)
= (10x^3 - x)/y
at (-1,2)
dy/dx = (-10+1)/2
= -9/2

so the tangent equation is 9x + 2y = c
with (-1,2) lying on it, so
-9 + 4 = c = -5

the tangent equation is 9x + 2y = -5

check:
http://www.wolframalpha.com/input/?i=y%5E2%3D5x%5E4-x%5E2+%2C+9x+%2B+2y+%3D+-5+%2C+from+-2+to+0

Answer 2

nope that is incorrect
the tangent equation is -9/2x-5/2

Answer 3

First of all -9/2x-5/2 is NOT an equation

if you meant y = -(9/2)x - 5/2
then by golly, that is what my equation is

9x + 2y = -5
2y = -9x - 5
divide each term by 2

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

My answer is correct.

Answer 4

well the online homework system didn't take it as a complete equation because the y= part was already outside of the answer box

Answer 5

well, there you go, how about that

Answer 6

It worked for mine, thanks Reiny.