So I was wondering if there was a formula for the point(s) of intersection between a parabola and a normal from a point on the parabola
1. Parabola: \(f(x) = y = ax^2+bx+c \)
2. Normal from a point on the parabola: \(n(x) = y = - \frac{1}{f'(x)}\cdot ( x - x_a ) + f(x_a)\)
3. \(f'(x_a) = 2ax_a+b \\y_a = f(x_a) = ax_a^2+bx_a+c\)
4. Intersection between a parabola and a normal:
\(\small{ \begin{array}{rcll} ax^2+bx+c &=& - \frac{1}{f'(x)}\cdot ( x - x_a ) + y_a \qquad | \qquad \cdot f'(x_a)\\ ax^2\cdot f'(x_a) + b\cdot f'(x_a)+c\cdot f'(x_a) &=& x_a -x + y_a \cdot f'(x_a) \\ \dots \\ x^2 \cdot [ a \cdot f'(x_a) ] +x \cdot [ 1+b\cdot f'(x_a) ]+ f'(x_a)\cdot (c-y_a) - x_a &=& 0 \\ \dots \\ \end{array} }\)
\(\boxed{~ \begin{array}{rcll} x_{1,2} &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] \pm \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\\\ && y_a = a\cdot x_a^2+b\cdot x_a+c \\\\ && f'(x_a) = 2a\cdot x_a +b \end{array} ~}\)
5. Example:
\(a=\frac12 \quad b= 0 \quad c = 0 \quad x_a = 4\)
Parabola: \(f(x) = \frac12x^2\)
\(f(x_a) = f(4) = y_a = \frac12 \cdot 4^2 = 8\\ f'(x_a) = f'(4) = 2\cdot \frac12 \cdot 4 + 0 = 4\)
\(\begin{array}{rcll} x_{1,2} &=& \dfrac{ -[~ 1+0 ~] \pm \sqrt{ 1+ [~ 4 ~]^2 \cdot [~ 2+ 0^2 -4\cdot \frac12\cdot(0-8) ~] } } { 2\cdot \frac12 \cdot 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 2 -4\cdot \frac12\cdot(-8) ~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 2+ 16~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 18~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm 17 } { 4 } \\\\ x_1 &=& \dfrac{ -1 + 17 } { 4 } \\ x_1 =x_a &=& 4\\\\ x_2 &=& \dfrac{ -1 - 17 } { 4 } \\ x_2 = x_b &=& -4.5 \\ y_2 = y_b &=& \frac12 \cdot x_b^2 \\ y_b &=& \frac12 \cdot (-4.5)^2 \\ y_b &=& 10.125 \end{array}\)
