Tschirnhausen Curve: Finding Tangents at a Given Point

[tc903]

Are the given families orthogonal trajectories of each other?

$$ {x}^{2}+{y}^{2} = ax $$
$$ {x}^{2}+{y}^{2} = by $$

I first started by finding them implicitly.

$$ \frac{2x+a}{2y} = y' $$
$$ \frac{2x+b}{2y} = y' $$

Then the problem wanted me to sketch my answer.

The Tschirnhausen, I solved. I just would like a better explanation.

$$ {y}^{2} = {x}^{3}+3{x}^{2} $$ Given $$\left(1,2\right)$$ find the tangent.

I found the tangent line. $$\frac{9}{4}x-\frac{1}{4}$$

Then the points where there was horizontal tangent. I take the terms in the numerator and set it equal to zero. $$ 3{x}^{2}+6x = 0 \implies 3x\left(x+2\right)=0 \implies x = 0, -2$$ Then I would substitute those values in for $$x$$ as my answers, but that is wrong, at least when $$x = 0$$. My answer should be $$\left(-2,2\right) and \left(-2,-2\right)$$. I was just wondering what I am missing.

 

[johng1]

Hi,
The short answer to your question about the Tshirnhausen curve is that when finding the derivative by implicit differentiation, you must not, as always, divide by 0. So if y is 0 your expression for y' is not right. The longer answer follows: