Question

Let y^2=4ax be a parallelogram and P be a point on it. Let the normal P intersects the x- axis at Q. Draw a line at Q perpendicular to the above normal. Show that this line intersects the parabola y^2+4a(x-2a)=0

Answer 1

Definition. The normal to the graph of a function at a point is a straight line passing through a given point perpendicular to the tangent to the graph of the function at this point (it is clear that a tangent must exist).

(more information : https://en.wikipedia.org/wiki/Normal_(geometry))

Definition. In geomerty, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point.

(more information : https://en.wikipedia.org/wiki/Tangent)

Usually, in theory we mean functions of the form $y=y(x)$ , then the tangent equation has the form

$$p.P(x_0,y(x_0))\longrightarrow y_{tan}(x)=y(x_0)+y'_x(x_0)\cdot(x-x_0)$$

But in our case, the function has the form $x=x(y)$ , so we will slightly alter the tangent equation

$$p.P(x(y_0),y_0)\longrightarrow x_{tan}(y)=x(y_0)+x'_y(y_0)\cdot(y-y_0)$$

1 STEP : Find the equation of the tangent and normal to the graph of the function $y^2=4ax$ at the point $p.P\left(x(y_0),y_0\right)$ :

$$y^2=4ax\to x(y)=\frac{y^2}{4a}\to x'_y=\frac{2y}{4a}=\frac{y}{2a}\to\\[0.3cm] x_{tan}(y)=x(y_0)+x'_y(y_0)\cdot(y-y_0)=\frac{y_0^2}{4a}+\frac{y_0}{2a}\cdot(y-y_0)$$

We are most interested in the slope $m$ , since we know that the angular coefficients of perpendicular lines are in the ratio :

$$\boxed{m_1\cdot m_2=-1}$$

In our case,

$$x_{tan}(y)=\frac{y_0^2}{4a}+\frac{y_0}{2a}\cdot(y-y_0)\to m_{tan}=\frac{y_0}{2a}\to\\[0.3cm] m_{tan}\cdot m_{norm}=-1\to m_{norm}=\frac{-1}{m_{tan}}=\frac{-1}{\frac{y_0}{2a}}=-\frac{2a}{y_0}\\[0.3cm] \text{The normal equation has the form} : x_{nomr}(y)=m_{nomr}y+b\\[0.3cm] x_{norm}(y)=-\frac{2a}{y_0}\cdot y+b$$

To find the coefficient "$b$ " we know that the normal also passes through that $p.P(x(y_0),y_0)$ , so its coordinates also satisfy the equation :

$$\frac{y_0^2}{4a}=-\frac{2a}{y_0}\cdot y_0+b\to\boxed{b=\frac{y_0^2}{4a}+2a}\\[0.3cm] \boxed{p.P(x(y_0),y_0)\longrightarrow x_{norm}(y)=-\frac{2a}{y_0}\cdot y+\frac{y_0^2}{4a}+2a}$$

2 STEP : Find the coordinates of the point $p.Q(x_0,y_0)$ - the point of intersection of the found normal and the $Ox-$ axis.

As we know, all points on the $Ox-$ axis have the form $p.M(x,0)$ . Then,

$$p.Q(x,0)\to x=-\frac{2a}{y_0}\cdot 0+\frac{y_0^2}{4a}+2a\to x=\frac{y_0^2}{4a}+2a\\[0.3cm] \boxed{p.Q\left(\frac{y_0^2}{4a}+2a,0\right)}$$

3 STEP : Let's find the equation of a straight line that passes through the point $p.Q$ and is perpendicular to the previously found normal.

Since the specified line must be perpendicular to the normal, this line is parallel to the previously found tangent. This means that its slope is the same as that of the tangent!

$$x_3(y)=m_{tan}\cdot y+b=\frac{y_0}{2a}\cdot y+b$$

To find the coefficient "$b$ " we know that the normal also passes through that $p.Q(x_0,0)$ , so its coordinates also satisfy the equation :

$$\frac{y_0^2}{4a}+2a=\frac{y_0}{2a}\cdot 0+b\to\boxed{b=\frac{y_0^2}{4a}+2a}\\[0.3cm] \boxed{x_3(y)=\frac{y_0}{2a}\cdot y+\frac{y_0^2}{4a}+2a}$$

4 STEP : Let us show that this line $x_3(y)$ intersects the parabola $y^2+4a(x-2a)=0$ .

Algebraically, this means that the system has a solution

$$\left\{\begin{array}{l} y^2+4a(x-2a)=0\\[0.3cm] x=\displaystyle\frac{y_0}{2a}\cdot y+ \displaystyle\frac{y_0^2}{4a}+2a\end{array}\right.\to y^2+4a\cdot\left(\displaystyle\frac{y_0}{2a}\cdot y+ \displaystyle\frac{y_0^2}{4a}+2a-2a\right)=0\\[0.3cm] y^2+\frac{4a}{1}\cdot\frac{y_0}{2a}\cdot\frac{y}{1}+\frac{4a}{1}\cdot\frac{y_0^2}{4a}=0\\[0.3cm] y^2+2yy_0+y_0^2=\left(y+y_0\right)^2=0\to y+y_0=0\to\boxed{y=-y_0}\\[0.3cm] x=\displaystyle\frac{y_0}{2a}\cdot (-y_0)+\displaystyle\frac{y_0^2}{4a}+2a=\frac{-y_0^2}{2a}+\frac{y_0^2}{4a}+2a=2a-\frac{y_0^2}{4a}\\[0.3cm] \boxed{p.M\left(2a-\frac{y_0^2}{4a},-y_0\right)-\text{solution of the specified system}}$$

Q.E.D.

Note : I will illustrate this task using the example of a parabola $y^2=4x(a=1)$