So we are doing parametric equations in class, and there are various fomulas for finding a number of things out about a parabola including the point of intersection of normals, of tangents, the gradients and equations of normals and tangents, the equation of a chord etc...
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. It makes sense that there should be, and I can solve it by just simultaneously solving the equations of the normal and parabola, but my varied attempts at generating a fomula have failed.
Here is what my formula ended up looking like, without any simplifications :
px^2 - 4ax - 4a^2p*(2 + p^2) = 0
But it didn't work.
Thanks in advance for any help :)
+26400 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: \(f(x) = \frac12x^2\)
\(f(x_a) = f(4) = y_a = \frac12 \cdot 4^2 = 8\\ f'(x_a) = f'(4) = 2\cdot \frac12 \cdot 4 + 0 = 4\)
\(\begin{array}{rcll} x_{1,2} &=& \dfrac{ -[~ 1+0 ~] \pm \sqrt{ 1+ [~ 4 ~]^2 \cdot [~ 2+ 0^2 -4\cdot \frac12\cdot(0-8) ~] } } { 2\cdot \frac12 \cdot 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 2 -4\cdot \frac12\cdot(-8) ~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 2+ 16~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm \sqrt{ 1+ 16 \cdot [~ 18~] } } { 4 } \\ x_{1,2} &=& \dfrac{ -1 \pm 17 } { 4 } \\\\ x_1 &=& \dfrac{ -1 + 17 } { 4 } \\ x_1 =x_a &=& 4\\\\ x_2 &=& \dfrac{ -1 - 17 } { 4 } \\ x_2 = x_b &=& -4.5 \\ y_2 = y_b &=& \frac12 \cdot x_b^2 \\ y_b &=& \frac12 \cdot (-4.5)^2 \\ y_b &=& 10.125 \end{array}\)
+26400 continuation
\(\boxed{~ \begin{array}{rcll} x_{1,2} &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] \pm \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\\\ && y_a = a\cdot x_a^2+b\cdot x_a+c \\\\ && f'(x_a) = 2a\cdot x_a +b \\\\ x_1 + x_2 &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] + \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\ && +\dfrac{ -[~ 1+b\cdot f'(x_a) ~] - \sqrt{ 1+ [~ f'(x_a) ~]^2 \cdot [~ 2+ b^2 -4\cdot a\cdot(c-y_a) ~] } } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& \dfrac{ -[~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } + \dfrac{ -[~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& \dfrac{ -2\cdot [~ 1+b\cdot f'(x_a) ~] } { 2\cdot a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& -\dfrac{ [~ 1+b\cdot f'(x_a) ~] } { a \cdot f'(x_a) } \\\\ x_1 + x_2 &=& -\frac{1}{a} \cdot \left[ \dfrac{ 1 } { f'(x_a) } + b \right] \qquad | \qquad f'(x_a) = 2ax_a+b \\\\ x_1 + x_2 &=& -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \qquad | \qquad x_1 = x_a \qquad x_2 = x_b \\\\ x_a + x_b &=& -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ x_b &=& -x_a -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ \end{array} ~}\)
conclusion
\(\boxed{~ \begin{array}{lcll} \text{Parabola } \quad y = ax^2+bx+c \qquad \text{ and we have } \quad a,b,c \text{ and } x_a \\\\ \Rightarrow x_b = -x_a -\frac{1}{a} \cdot \left( \dfrac{ 1 } { 2ax_a+b } + b \right) \\\\ \Rightarrow y_b = ax_b^2+bx_b+c \\\\ \end{array} ~}\)
Example:
\(\begin{array}{rcll} \text{Parabola } \quad y &=& \frac12 x^2 \qquad a=\frac12 \qquad b=0 \qquad c=0 \qquad x_a=4 \\\\ x_b &=& -4 -\frac{1}{\frac12} \cdot \left( \dfrac{ 1 } { 2\cdot \frac12 \cdot 4+0 } + 0 \right) \\\\ x_b &=& -4 -\frac{1}{\frac12} \cdot \left( \dfrac{ 1 } { 4 } \right) \\\\ x_b &=& -4 - \frac24 \\\\ x_b &=& -4 - \frac12 \\\\ \mathbf{x_b} & \mathbf{=} & \mathbf{-4.5} \\\\\\ y_b &=& \frac12 \cdot (-4.5)^2+0\cdot (-4.5)+0 \\\\ \mathbf{y_b} &\mathbf{=}& \mathbf{10.125} \\\\ \end{array}\)
