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 $b$…
Best Answer
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 $$(x-2)^2+y^2=9.$$
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 $$(x-2)^2+y^2=9.$$