A curve has the property that the normal line through any point on the curve passes through $(2,0)$. Find its equation.

Question

A curve has the property that the normal line through any point on the curve passes through $(2,0)$. If the curve contains the point $(2,3)$ find its equation.

My attempt:

Assume $(a,b)$ is a point on the function. The normal line has a slope $m=\frac{b}{a-2}$ and hence the tangent line has a slope $m_T = \frac{2-a}{b}$. Hence, this is the slope we can integrate to get the equation:

$$F(x) = \int \frac{2-a}{b}x dx = \frac{2-a}{2b}x^2+C$$

Since we know that $(2,3)$ is a point we can isolate for $C$ to get:

$$F(x) = \frac{2-a}{2b}x^2+\frac{3b-4+2a}{b}$$

I am getting stuck here because I am not sure how to get values for a $a$ and $b$...

Answer 1

A circle has the property that the normal through any point on the curve will pass through its center. As such, we can deduce that the center of the circle is $(2,0)$ and the circle contains the point $(2,3)$. Thus, we have the equation $$(x-2)^2+y^2=9.$$

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Taking an approach using calculus instead of geometry:

Define $-\dfrac{dx}{dy}$ as the slope of the normal to the curve $y(x)$. The equation of the normal to the curve that contains $(x_1,y_1)$ would be $$y-y_1=-\frac{dx}{dy}(x-x_1).$$ Since the normal must contain $(2,0)$, we can substitute and solve a differential equation: \begin{align} y-0&=-\frac{dx}{dy}(x-2)\\ y\,dy&=(2-x)\,dx\\ \int y\,dy&=\int2-x\,dx\\ \frac12y^2&=-\frac12(2-x)^2+C_1\\ (x-2)^2+y^2&=C_2 \end{align}

Since this curve must contain $(2,3)$, it follows that $C_2=9$. Therefore, the curve must have the equation $$(x-2)^2+y^2=9.$$

Answer 2

A somewhat more general proof, but one which is nevertheless in the spirit of our colleague Andrew Chin's answer, may be had using elementary differential geometry:

A Little Lemma: Let $\alpha(s)$ be an arc-length parametrized curve in $\Bbb R^2$ such that every normal line to $\alpha(s)$ passes through the point $C \in \Bbb R^2$. Then $\alpha(s)$ lies in a circle centered at $C$.

Proof of Little Lemma: Let $T(s)$ be the unit tangent vector field to $\alpha(s)$; that is

$T(s) = \dot \alpha(s); \tag 1$

by the first Frenet-Serret equation, $T$ is related to the curvature $\kappa(s)$ of $\alpha(s)$ via the formula

$\dot T(s) = \kappa(s) N(s), \tag 2$

where

$\kappa(s) > 0 \tag 3$

and $N(s)$ is the unit normal vector field to $\alpha(s)$. Then the line through $\alpha(s)$ in the direction $N(s)$ passes through $C$; if $l(s)$ denotes the signed length of the segment joining $\alpha(s)$ and $C$, we have

$\alpha(s) + l(s)N(s) = C, \tag 4$

holding for all $s$ in the domain of $\alpha(s)$. (Note that $l(s)$ is differentiable in $s$ since we may write (4) in the form

$l(s)N(s) = C - \alpha(s), \tag{4.1}$

from which

$l(s) = l(s)N(s) \cdot N(s) = (C - \alpha(s)) \cdot N(s), \tag{4.2}$

with the right-hand side differentiable in $s$.)

We differentiate equation (4) with respect to $s$, and obtain

$\dot \alpha(s) + \dot l(s) N(s) + l(s) \dot N(s) = 0, \tag 5$

since $C$ does not depend on $s$.

The companion equation to (2),

$\dot N(s) = -\kappa(s)T(s), \tag 6$

which may be derived starting from the orthogonality of $T$ and $N$,

$T(s) \cdot N(s) = 0, \tag 7$

and then differentiating:

$\dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag 8$

and substituting $\dot T(s)$ from (2):

$\kappa(s) N(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0; \tag 9$

since

$N(s) \cdot N(s) = 1, \tag{10}$

(9) may be written

$T(s) \cdot \dot N(s) = -\kappa(s); \tag{11}$

now, differentiating (10) yields

$\dot N(s) \cdot N(s) = 0, \tag{12}$

which shows that $\dot N(s)$ is parallel to $T(s)$; then in light of (11) we infer

$\dot N(s) = -\kappa(s)T(s), \tag{13}$

and substituting this and (1) into (5) we write

$T(s) + \dot l(s) N(s) - l(s) \kappa(s) T(s) = 0, \tag{14}$

i.e.,

$(1 - l(s) \kappa(s))T(s) + \dot l(s) N(s) = 0; \tag{15}$

the orthnormality of $T(s)$ and $N(s)$ now allows us to conclude that

$1 - l(s) \kappa(s) = 0, \tag{16}$

$\dot l(s) = 0, \tag{17}$

which show that both $l(s) = l$ and $\kappa(s) = \kappa$ are constant with

$\kappa = \dfrac{1}{l}, \tag{18}$

and hence in light of (3) that

$l > 0; \tag{18.1}$

finally, from (4)

$\vert \alpha(s) - C \vert = \vert l(s)N(s) \vert = l(s) \vert N(s) \vert = l, \tag{19}$

which shows that $\alpha(s)$ lies in the circle of radius $l$ centered at $C$. End: Proof of Litte Lemma.

Now taking

$C = (2, 0) \tag{20}$

and

$\alpha(s) = (2, 3) \tag{21}$

for some $s$, we find that

$l = 3, \tag{22}$

and thus see that $\alpha(s)$ lies in the circle of radius $3$ about $(2, 0)$, the same result as obtained by Andrew Chin in his answer. The equation of the circle in which $\alpha(s)$ lies is of course

$(x - 2)^2 + y^2 = 9. \tag{23}$