Question

The curve with equation y2 = 5x4 − x2 is called a kampyle of eudoxus. find an equation of the tangent line to this curve at the point (−1, 2).

Answer 1

To find the tangent line we will need the slope of the tangent at x=-1 (the x-coordinate of the point given). We find the slope by using the derivative of the curve.

Th curve given is $y^{2} =5 x^{4} - x^{2}$ which can be solved for y by taking the root of both sides. We obtain $y=(5 x^{4}- x^{2} ) ^{1/2}$

We find the derivative using the chain rule. Bring down the exponent, keep the expression in the parenthesis, raise it to 1/2 - 1 and then take the derivative of what is inside.

$y’=(1/2)(5 x^{4} - x^{2} )^{-1/2}(20 x^{3} -2x)$

Next we evaluate this expression for x=-1 and obtain:
$\frac{20(-1)^{3} -(2)(-1)}{2 \sqrt{5(-1 ^{4}-(-1 ^{2}) } }= \frac{-20+2}{2 \sqrt{5-1} }=-18/4=-9/2$

So we are looking for a line through (-1,2) with slope equal to -9/2. We use y=mx+b with m=-9/2, x=-1 and y=2 to find b.

$2=(-9/2)(-1)+b$
2-(9/2)=b
b=-5/2

So the tangent line is given by y=(-9/2)x+(-5/2)