# Tangent and Normal lines that pass through the origin

Let $\alpha$ be a regular curve in $\mathbb{R}^2$ and let all of its tangent lines pass through the origin. Also, let $\beta$ be a regular curve in $\mathbb{R}^2$ and let all of its normal lines pass through the origin.

How can I show that $\alpha$ is contained in a straight line through the origin and that $\beta$ is contained in a circle around the origin?

I know that the tangent line $\alpha(t)$ is the line that points in the direction of the tangent vector $T(t)$ and that the normal line of $\beta(t)$ is the line that points in the direction of $N(t)$, but how can I put this all together to complete the proof?

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The tangent line at the point $(t,\alpha(t))$ to the curve $\alpha$ passes through the origin and must have the equation $$y=\alpha'(t)x$$ As the point $(t,\alpha(t))$ lies on this line the curve $\alpha$ satisfies $$\alpha(t)=t\alpha'(t),~~~~~\forall t$$ Solve this to get the parametric equation of $\alpha$ in terms of $t$.

Similarly you can show that $\beta$ satisfies $$\beta(t)=\frac{t}{\beta'(t)}$$ and then solve this differential equation.

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Im not sure how you derived your equations. Where did you get $y=\alpha'(t)x$ and $\alpha(t)=t\alpha'(t),~~~~~\forall t$ from, also $\beta$? – Lays Sep 14 '13 at 3:45
The slope of the line tangent to the curve $\alpha$ at the point $t$ has to be $\alpha'(t)$. Now write down the equation of the line that passes through the origin and has slope $\alpha'(t)$. Now this line also passes through the point on the curve, $(t,\alpha(t))$, where it is tangent to. So if the equation of the straight line is $y=\alpha'(t)x$, then $\{x=t,y=\alpha(t)\}$ must satisfy this equation. – Abishanka Saha Sep 14 '13 at 3:54

Let $c$ be a regular parametrization of the curve $\beta$. At each point of the curve a unique normal line passing through it is well defined as we are in $\mathbb{R}^2$ and there is a tangency direction because of the regularity of the parametrization. As the normal line at $c(t)$ passes through the origin it has the direction $c(t)-0 = c(t)$. And then, we deduce $\langle \dot{c}(t) , c(t) \rangle = 0$ because of them being orthogonal.

Applying this relation we get:

$\frac{d}{dt} \| c(t)\|^2 = \frac{d}{dt} \langle c(t) , c(t) \rangle = 2\langle \dot{c}(t) , c(t) \rangle = 0$

As the reasoning works for every $t$ we have that $\| c(t)\|^2$ is constant as a function of $t$ and so is $\| c(t)\|$. Putting $R$ for this last constant we have that $\beta$ is contained in the circumference of center $0$ and radius $R$.

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Thank you Pip ! – Lays Sep 15 '13 at 8:07