Given the parabolas y²+4x-4 = 0 and 8x²-2x-3y-6 = 0, draw
Given the parabolas $y^{2}+4x-4 = 0$ and $8x^{2}-2x-3y-6 = 0$, draw the normals to the first at their points of intersection and show that three of these normals meet at a single point.
Explanation of Method
Four intersection points, four normals, and one meeting point for three of them.
Step 1: The intersection points
From the first curve, $x = 1-\dfrac{y^{2}}{4}$. Substituting into the second: \[8\left(1-\frac{y^{2}}{4}\right)^{2}-2\left(1-\frac{y^{2}}{4}\right)-3y-6 = \frac{y^{4}}{2}-\frac{7y^{2}}{2}-3y = 0,\]
\[ y^{4}-7y^{2}-6y = y\bigl(y^{3}-7y-6\bigr) = y\bigl(y+1\bigr)\bigl(y-3\bigr)\bigl(y+2\bigr) = 0. \tag{1} \]
Hence $y = 0,\,-1,\,3,\,-2$ and, from $x = 1-\tfrac{y^{2}}{4}$, \[P_{1} = \bigl(1,0\bigr), \quad P_{2} = \left(\frac34,-1\right), \quad P_{3} = \left(-\frac54,3\right), \quad P_{4} = \bigl(0,-2\bigr). \tag{2}\]
Step 2: The slopes of the normals
Differentiating $y^{2}+4x-4 = 0$ implicitly, $2yy'+4 = 0$, so $y' = -\dfrac2y$ and the normal has slope
\[ m = -\frac{1}{y'} = \frac{y_{0}}{2} \qquad \bigl(y_{0}\neq0\bigr). \tag{3} \]
At $P_{1}$ the tangent is vertical, so the normal there is the horizontal line $y = 0$.
Step 3: The four normals
\[ n_{1}:\ y = 0; \qquad n_{2}:\ y = -\frac{x}{2}-\frac58; \qquad n_{3}:\ y = \frac{3x}{2}+\frac{39}{8}; \qquad n_{4}:\ y = -x-2. \]
Step 4: Three of them concur
Intersecting $n_{2}$ and $n_{4}$: \[-\frac{x}{2}-\frac58 = -x-2 \;\Longrightarrow\; \frac{x}{2} = -\frac{11}{8} \;\Longrightarrow\; x = -\frac{11}{4}, \qquad y = \frac{11}{4}-2 = \frac34.\]
And $n_{3}$ passes through that point: \[\frac32\left(-\frac{11}{4}\right)+\frac{39}{8} = -\frac{33}{8}+\frac{39}{8} = \frac34 \;\;\checkmark\]
The fourth normal $n_{1}$ does not: it is the line $y = 0$, while the common point has $y = \tfrac34$.
Step 5: Verification
\[\text{All four points satisfy the second equation: } 8\cdot\tfrac9{16}-\tfrac32+3-6 = 0 \ \text{at } P_{2};\ 8\cdot\tfrac{25}{16}+\tfrac52-9-6 = 0 \ \text{at } P_{3} \;\;\checkmark\]
\[P_{1}:\ 8-2-0-6 = 0; \qquad P_{4}:\ 0-0+6-6 = 0 \;\;\checkmark\]
\[\text{Slopes of } n_{2},n_{3},n_{4}:\ -\tfrac12,\ \tfrac32,\ -1 = \tfrac{y_{0}}{2} \text{ for } y_{0} = -1,3,-2 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,n_{2},\ n_{3},\ n_{4} \ \text{meet at } S = \left(-\frac{11}{4},\;\frac34\right); \qquad n_{1}:\ y = 0 \ \text{does not pass through } S\,}\]
Remark
Remark. The concurrence is no accident, and the algebra shows why. Asking which normals of $y^{2} = 4-4x$ pass through $S = \left(-\tfrac{11}{4},\tfrac34\right)$ gives $\tfrac34-y_{0} = \tfrac{y_{0}}{2}\left(-\tfrac{11}{4}-1+\tfrac{y_{0}^{2}}{4}\right)$, that is $y_{0}^{3}-7y_{0}-6 = 0$ — precisely the cubic factor that appeared in (1). The second parabola has been chosen so that its intersections with the first carry exactly those three ordinates, plus the extra root $y = 0$ whose normal, $n_{1}$, is the odd one out. A parabola admits at most three normals through any one point, so three is the most that could have been asked for.