Show that the normals to the curve $y=4x^2$ from points the same distance on either side of the y-axis intersect on the y-axis.
My attempt, I differentiated so it becomes $8x$, but I don't understand the question. Can anyone explain it to me ? Thanks in advance:
The distance from a point $P=(a,4a^2)$ of the curve to the y-axis is $d=|a|$, so two points of the curve has the same distance to y-axis then they are $(a,4a^2)$ and $(-a,4a^2)$.
The normal has slope $m=-\frac{1}{8x_0}$ and then it has the form
$$y=-\frac{1}{8x_0}x+k$$
If it goes trough $(a,4a^2)$ we get
$$y_1=-\frac{1}{8a}x+\frac{32a^2+1}{8a}$$
If it goes through $(-a,4a^2)$ we get
$$y_2=\frac{1}{8a}x+\frac{32a^2+1}{8a}$$
So the point
$$\left(0,\frac{32a^2+1}{8a}\right)$$
is the intersection
That parabola is symmetric with respect to $y$-axis, hence the normal lines issued from two symmetric points (as in your case) are corresponding one another through a reflection across $y$-axis. Unless they are both parallel (which is not possible if they are issued from different points) they must therefore meet on the $y$-axis.
The normals or tangents to any continuous/smooth curve having an axis of symmetry $A$ at oppositely situated points (mirrored about $A$) must necessarily meet on $A$.