tc903
- 19
- 0
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.
$$ {x}^{2}+{y}^{2} = ax $$
$$ {x}^{2}+{y}^{2} = by $$
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.