On the parabola y²=4kx take A(k alpha²,2k alpha) and B(k

On the parabola $y^2=4kx$ take $A(k\alpha^2,2k\alpha)$ and $B(k\beta^2,2k\beta)$, $\alpha\neq\beta$. 1° Find the tangent and normal at $A$. 2° Let $S$ be the intersection of the tangents at $A,B$ and $M$ the intersection of the normals at $A,B$. Find the locus of $M$ as $S$ moves along a line parallel to the axis of the parabola.

Step 1: Tangent and normal at $A$

Implicit differentiation of $y^2=4kx$ gives $y'=2k/y$; at $A$, $y'|_A=2k/(2k\alpha)=1/\alpha$. The tangent, $y-2k\alpha=\frac1\alpha(x-k\alpha^2)$, simplifies to

\[ x - \alpha y + k\alpha^{2} = 0. \tag{1} \]

The normal, with slope $-\alpha$, simplifies to

\[ \alpha x + y - k\alpha(2+\alpha^{2}) = 0. \tag{2} \]

Step 2: The tangent-intersection $S$

Subtracting the tangent at $B$ (equation (1) with $\beta$ in place of $\alpha$) from (1) and factoring out $\beta-\alpha\neq0$: \[S = \bigl(k\alpha\beta,\ k(\alpha+\beta)\bigr). \tag{3}\]

Step 3: The condition on $S$

The axis of the parabola is the $x$-axis, so a line parallel to it is $y=c$; requiring $S$ to lie on it means $y_S=k(\alpha+\beta)=c$, i.e.

\[ \alpha+\beta = \frac{c}{k} =: m = \text{const.} \tag{4} \]

The product $p=\alpha\beta$ is left free as $A,B$ slide over the curve subject to (4).

Step 4: The normal-intersection $M$

Subtracting the normal at $B$ from (2) and factoring out $\beta-\alpha$: \[x_M = k\bigl(2+\alpha^2+\alpha\beta+\beta^2\bigr) = k(2+m^2-p), \tag{5}\] using $\alpha^2+\beta^2=m^2-2p$. Substituting $x_M$ back into (2) and simplifying ($\alpha\beta=p$): \[y_M = -k\alpha\beta(\alpha+\beta) = -kpm. \tag{6}\]

Step 5: Eliminate the free parameter $p$

From (6), $p=-y_M/(km)$; substituting into (5): \[x_M = k(2+m^2) + \frac{y_M}{m}. \tag{7}\]

Replacing $m=c/k$ and writing $(x,y)$ for the running point $M$: \[x - \frac{k}{c}\,y = 2k+\frac{c^2}{k}. \tag{8}\]

Step 6: Verification

\[k=1,\ c=2,\ m=2\text{: for }\alpha=0.6,1.0,1.7\ (\beta=m-\alpha)\text{, every resulting }M\text{ satisfies }x-y=2+2=4\text{ to machine precision}\ \checkmark\]